BINOMIAL THEOREM - 1 (Principle and simple applications)
Binomial Theorem for Positive Integral Index
If $\mathrm{x}$ and $\mathrm{y}$ are real, then for all $\mathrm{n} \in \mathrm{N}$
$ \begin{aligned} & (x+y)^{n}={ }^{n} C _{0} x^{n}+{ }^{n} C _{1} x^{n-1} y+{ }^{n} C _{2} x^{n-2} y^{2}+\ldots \ldots .+{ }^{n} C _{r} x^{n-r} y^{r}+\ldots .+{ }^{n} C _{n-1} x^{1} y^{n-1}+{ }^{n} C _{n} y^{n} \\ & =\quad \sum _{r=0}^{n}{ }^{n} C _{r} x^{n-r} y^{r} \\ & (x-y)^{n}={ }^{n} C _{0} x^{n}-{ }^{n} C _{1} x^{n-1} y^{1}+{ }^{n} C _{2} x^{n-2} y^{2} \ldots \ldots .{ }^{n} C _{r}(-1)^{r} x^{n-r} y^{r}+\ldots .{ }^{n} C _{n-1}(-1)^{n-1} x^{1} y^{n-1}+{ }^{n} C _{n}(-1)^{n} y^{n} \\ & (x+y)^{n}+(x-y)^{n}=2\left\{{ }^{n} C _{0} x^{n}+{ }^{n} C _{2} x^{n-2} y^{2}+{ }^{n} C _{4} x^{n-4} y^{4}+\ldots \ldots . .\right. \\ & (x+y)^{n}-(x-y)^{n}=2\left\{{ }^{n} C _{1} x^{n-1} a^{1}+{ }^{n} C _{3} x^{n-3} a^{3}+{ }^{n} C _{5} x^{n-5} y^{5}+\ldots \ldots . .\right] \\ & (1+x)^{n}={ }^{n} C _{0}+{ }^{n} C _{1} x+{ }^{n} C _{2} x^{2}+\ldots \ldots \ldots{ }^{n} C _{r} x^{r}+\ldots \ldots .{ }^{n} C _{n} x^{n} \\ & (1-x)^{n}={ }^{n} C _{0}-{ }^{n} C _{1} x+{ }^{n} C _{2} x^{2}-\ldots \ldots \ldots .{ }^{n} C _{r}(-1)^{r} x^{r}+\ldots \ldots+{ }^{n} C _{n}(-1)^{n} x^{n} \end{aligned} $
Properties of Binomial Expansion
(i). The number of terms in the expansion of $(x+y)^{n}$ where $n \in \mathrm{N}$ is $(\mathrm{n}+1)$.
(ii). The sum of exponents of $x & y$ in $(x+y)^{n}$ is equal to $n$, the index of the expansion.
(iii). Since ${ }^{n} \mathrm{C} _{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}-\mathrm{r}} ; \mathrm{r}=0,1,2, \ldots \ldots . \mathrm{n}$, the binomial coefficients equidistant from the begin ning and the end are equal.
i.e. ${ }^{n} C _{0}={ }^{n} C _{n},{ }^{n} C _{1}={ }^{n} C _{n-1}$ and so on.
(iv). The general term is the expansion of $(\mathrm{x}+\mathrm{y})^{\mathrm{n}}$ is given by
$ T _{r+1}={ }^{n} C _{r} x^{n-r} y^{r} $
(v). Coefficient of $(r+1)^{\text {th }}$ term in the expansion of $(1+x)^{n}$ is ${ }^{n} C _{r}=$ coefficient of $x^{r}$.
(vi). If $\mathrm{n}$ is odd, then $\left\{(\mathrm{x}+\mathrm{y})^{\mathrm{n}}+(\mathrm{x}-\mathrm{y})^{\mathrm{n}}\right\}$ and $\left\{(\mathrm{x}+\mathrm{y})^{\mathrm{n}}-(\mathrm{x}-\mathrm{y})^{\mathrm{n}}\right\}$ have same number of terms equal to $\left(\frac{\mathrm{n}+1}{2}\right)$.
If $n$ is even, then
$\left\{(\mathrm{x}+\mathrm{y})^{\mathrm{n}}+(\mathrm{x}-\mathrm{y})^{\mathrm{n}}\right\}$ has $\left(\frac{\mathrm{n}}{2}+1\right)$ terms and
$\left\{(\mathrm{x}+\mathrm{y})^{\mathrm{n}}-(\mathrm{x}-\mathrm{y})^{\mathrm{n}}\right\}$ has $\left(\frac{\mathrm{n}}{2}\right)$ terms.
(vii). Middle term If $\mathrm{n}$ is even then in the expansion of $(x+y)^{n},\left(\frac{n}{2}+1\right)$ th terms is the middle term.
If $\mathrm{n}$ is odd natural number, then $\left(\frac{\mathrm{n}+1}{2}\right)$ th and $\left(\frac{\mathrm{n}+3}{2}\right)$ th are the middle terms in the expansion of $(x+y)^{n}$.
(viii). Let $\mathrm{S}=(\mathrm{x}+\mathrm{y})^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C} _{0} \mathrm{x}^{\mathrm{n}}+{ }^{n} \mathrm{C} _{1} \mathrm{x}^{\mathrm{n}-1} \mathrm{y}+\ldots \ldots .+{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}-1} \mathrm{xy}^{\mathrm{n}-1}+{ }^{n} \mathrm{C} _{\mathrm{n}} \mathrm{y}^{\mathrm{n}}$ where $\mathrm{n} \in \mathrm{N}$
$= \sum _{\mathrm{r}=0}^{\mathrm{n}} \mathrm{C} _{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}} \mathrm{y}^{\mathrm{r}}$
Replacing $\mathrm{r}$ by $\mathrm{n}-\mathrm{r}$ we get,
$S=\sum _{r=0}^{n}{ }^{n} C _{n-r} x^{r} y^{n-r}$
$={ }^{n} C _{n} y^{n}+{ }^{n} C _{n-1} y^{n-1} x+\ldots . .+{ }^{n} C _{n-1} y^{n-1}+{ }^{n} C _{n} x^{n}$.
i.e. By replacing $\mathrm{r}$ by $\mathrm{n}-\mathrm{r}$, we are writing the binomial expansion in the reverse order
Properties of Binomial Coefficient
(i). Sum of two binomial coefficients, ${ }^{n} C _{r}+{ }^{n} C _{r-1}={ }^{n+1} C _{r}$
(ii). ${ }^{n} C _{r}=\frac{n}{r}{ }^{n-1} C _{r-1}$
(iii). $\frac{{ }^{n} C _{r}}{{ }^{n} C _{r-1}}=\frac{n-r+1}{r}$
(iv). ${ }^{n} \mathrm{C} _{\mathrm{r}}={ }^{n} \mathrm{C} _{\mathrm{s}} \Rightarrow$ either $\mathrm{r}=\mathrm{s}$ or $\mathrm{r}+\mathrm{s}=\mathrm{n}$
Multinomial Theorem (For a positive integral index)
$\left(\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots \ldots .+\mathrm{xk}\right)^{\mathrm{n}}=\frac{\mathrm{n} !}{\mathrm{n} _{1} ! \mathrm{n} _{2} ! \ldots . . \mathrm{nk} !} \mathrm{x} _{1}^{\mathrm{n} 1} \mathrm{x} _{2}^{\mathrm{n} 2} \ldots \ldots \mathrm{x} _{\mathrm{k}}^{\mathrm{nk}}$,
Where $\mathrm{n} _{1}+\mathrm{n} _{2}+\ldots \ldots+\mathrm{n} _{\mathrm{k}}=\mathrm{n}$ and $0 \leq \mathrm{n} _{1}, \mathrm{n} _{2}, \ldots \ldots . \mathrm{n} _{\mathrm{k}} \leq \mathrm{n}$
- The greatest coefficient in this expansion is $\frac{\mathrm{n} !}{(\mathrm{q} !)^{k-\mathrm{r}}((\mathrm{q}+1) !)^{\mathrm{r}}}$ where $\mathrm{q}$ is the quotient and $\mathrm{r}$ is the remainder when $\mathrm{n}$ is divided by $\mathrm{k}$.
$\quad \quad $ Eg. Find the greatest coefficient in $(x+y+z+w)^{15}$
$\quad \quad $ $\mathrm{n}=15, \mathrm{k}=4$ we have $15=4 \times 3+3$ i.e. $\mathrm{q}=3, \mathrm{r}=3$ greatest coefficient $=\frac{15 !}{(3 !)^{1}(4 !)^{3}}$
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Number of distinct terms in the expansion is ${ }^{n+k-1} C _{k-1}$ (Total number of terms).
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Number of positive integer solutions of $\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots+\mathrm{x} _{\mathrm{k}}=\mathrm{n}$ is ${ }^{\mathrm{n}-1} \mathrm{C} _{\mathrm{k}-1}$.
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Number of non negative integer solutions of $x _{1}+x _{2}+\ldots+x _{k}=n$ is ${ }^{n+k-1} C _{k-1}$.
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Sum of all the coefficients is obtained by setting $\mathrm{x} _{1}=\mathrm{x} _{2}=\ldots \ldots \mathrm{x} _{\mathrm{k}}=1$.
Greatest Coefficient and Greatest term
Consider the binomial expansion of $(x+y)^{n}$. where $n \in W$. For a given value of $n$,
Maximum value of ${ }^{n} \mathrm{C} _{\mathrm{r}}$ is ${ }^{n} \mathrm{C} _{\mathrm{n} 2}$ if $\mathrm{n}$ is even
Maximum value of ${ }^{n} C _{r}$ is ${ }^{n} C _{\frac{n-1}{2}}={ }^{n} C _{\frac{n+1}{2}}$ if $n$ is od(d).
To find the greatest term in the expansion of $(x+y)^{n}$,
(i) Find $m=\frac{n+1}{1+\frac{x}{y}}$
(ii) If $\mathrm{m}$ is an integer we have $\mathrm{m}^{\text {th }}$ and $(\mathrm{m}+1)^{\text {th }}$ terms as greatest terms.
(iii) If $\mathrm{m}$ is not an integer, then $([\mathrm{m}]+1)$ th term is the greatest term where [.] denote the greatest integer $\leq \mathrm{m}$.
Divisibility Problems
From the expansion
$(1+\mathrm{a})^{\mathrm{n}}=1+{ }^{\mathrm{n}} \mathrm{C} _{1} \mathrm{a}+{ }^{\mathrm{n}} \mathrm{C} _{2} \mathrm{a}^{2}+\ldots . .+{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}} \mathrm{a}^{\mathrm{n}}$, we can see that
(i) $\quad(1+a)^{n}-1$ is a mutliple of $a=M(a)$
(ii) $\quad (1+\mathrm{a})^{\mathrm{n}}-1-$ na is a mutliple of $\mathrm{a}^{2}=\mathrm{M}\left(\mathrm{a}^{2}\right)$
(iii) $\quad (1+\mathrm{a})^{\mathrm{n}}-1-\mathrm{na}-\frac{\mathrm{n}(\mathrm{n}-1)}{2} \mathrm{a}^{2}$ is a mutliple of $\mathrm{a}^{3}$ and so on.
For example
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$(1+8)^{50}-1=9^{\mathrm{n}}-1$ is $\mathrm{M}(8)$
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$(1+8)^{50}-1-50 \times 8=9^{\mathrm{n}}-399$ is $\mathrm{M}\left(8^{2}\right)=\mathrm{M}(64)$
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$(1+8)^{50}-1-50 \times 8-\frac{50 \times 49}{2} 8^{2}=\mathrm{M}\left(8^{3}\right)=\mathrm{M}(512)$ and soon
Binomial Theorem for any index (for negative or fractional index)
$\quad$ If $n \varepsilon Q$, then $(1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\ldots . . \infty$ provided $|x|<1$.
- For any index $n$, the general term in the expansion of
i. $\quad(1+x)^{n}$ is $T _{r+1}=\frac{n(n-1) \ldots \ldots . .(n-r+1)}{r !} x^{r}$
ii. $\quad(1+\mathrm{x})^{-\mathrm{n}}$ is $\mathrm{T} _{\mathrm{r}+1}=\frac{(-1)^{\mathrm{r}} \mathrm{n}(\mathrm{n}+1) \ldots \ldots \ldots .(\mathrm{n}+\mathrm{r}-1)}{\mathrm{r} !} \mathrm{x}^{\mathrm{r}}$
iii. $\quad(1-\mathrm{x})^{\mathrm{n}}$ is $\mathrm{T} _{\mathrm{r}+1}=\frac{(-1)^{\mathrm{r}} \mathrm{n}(\mathrm{n}-1) \ldots \ldots \ldots .(\mathrm{n}-\mathrm{r}+1)}{\mathrm{r} !} \mathrm{x}^{\mathrm{r}}$
iv. $\quad(1-\mathrm{x})^{-\mathrm{n}}$ is $\mathrm{T} _{\mathrm{r}+1}=\frac{\mathrm{n}(\mathrm{n}+1) \ldots \ldots \ldots .(\mathrm{n}+\mathrm{r}-1)}{\mathrm{r} !} \mathrm{x}^{\mathrm{r}}$
- The following expansions should be remembered (for $|\mathrm{x}|<1$ ).
i. $\quad (1+x)^{-1}=1-x+x^{2}-x^{3}+\ldots \ldots \ldots \infty$
ii. $\quad (1-x)^{-1}=1+x+x^{2}+x^{3}+$ $\ldots\ldots \ldots\infty$
iii. $\quad (1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+$ $\ldots\ldots \ldots\infty$
iv. $\quad (1-x)^{-2}=1+2 x+3 x^{2}+4 x^{3}+$ $\ldots\ldots \ldots\infty$
- Note : The expansion in ascending powers of $x$ is valid if $x$ is small. If $x$ is large (i.e. $|x|>1)$, then we may find it convenient to expand in powers of $\frac{1}{x}$, which then will be small.
Exponential series
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$ \mathrm{e}^{\mathrm{x}}=1+\frac{\mathrm{x}}{1 !}+\frac{\mathrm{x}^{2}}{2 !}+\frac{\mathrm{x}^{3}}{3 !}+\ldots . . \infty$
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$ \mathrm{e}=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots . \infty(\mathrm{e} \simeq 2.72)$
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$ \mathrm{e}+\mathrm{e}^{-1}=2\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots . \infty\right)$
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$ \mathrm{e}-\mathrm{e}^{-1}=2\left(\frac{1}{1 !}+\frac{1}{3 !}+\frac{1}{5 !}+\frac{1}{7 !}+\ldots . \infty\right)$
Logarithmic series
For $-1<\mathrm{x} \leq 1$
$\log _{e}(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots . . \infty$
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$ \log _{\mathrm{e}}(1-\mathrm{x})=-\mathrm{x}-\frac{\mathrm{x}^{2}}{2}-\frac{\mathrm{x}^{3}}{3}-\frac{\mathrm{x}^{4}}{4}+\ldots . \infty,-1 \leq \mathrm{x}<1$
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$ \log \left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)=2\left(\mathrm{x}+\frac{\mathrm{x}^{3}}{3}+\frac{\mathrm{x}^{5}}{5}+\ldots . \infty\right),-1<\mathrm{x}<1$
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$\log _{\mathrm{e}} 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots . . \infty \approx 0.693$
Solved examples
1. If $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2}, \ldots . . \mathrm{Cn}$ denote the coefficients in the binomial expansion of $(1+\mathrm{x})^{\mathrm{n}}$, prove that :
$ \mathrm{C} _{1}+2 \mathrm{C} _{2}+3 \mathrm{C} _{3}+\ldots . .+\mathrm{nC}=\mathrm{n} \cdot 2^{\mathrm{n}-1} $
i.e. $ \sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r} \cdot \mathrm{C} _{\mathrm{r}}=\mathrm{n} \cdot 2^{\mathrm{n}-1}$
Show Answer
Solution :
We have
$\mathrm{C} _{1}+2 \mathrm{C} _{2}+3 . \mathrm{C} _{3}+\ldots .+\mathrm{nC} _{\mathrm{n}}$
$=\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r} \cdot \mathrm{C} _{\mathrm{r}}$
$=\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{\mathrm{n}} \mathrm{C} _{\mathrm{r}} \quad\left[\because \mathrm{C} _{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}\right]$
$=\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r} \cdot \frac{\mathrm{n}}{\mathrm{r}}{ }^{\mathrm{n}-1} \mathrm{C} _{\mathrm{r}-1} \quad\left[\because{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}=\frac{\mathrm{n}}{\mathrm{r}} .{ }^{\mathrm{n}-1} \mathrm{C} _{\mathrm{r}-1}\right]$
$=\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{n} _{\mathrm{r}-1}$
$=\mathrm{n}\left({ }^{(n-1} C _{0}+{ }^{n-1} C _{1}+\ldots .+{ }^{n-1} C _{n-1}\right)=\mathrm{n}(1+1)^{\mathrm{n}-1}[\because \mathrm{x}=1]$
$=n \cdot 2^{\mathrm{n}-1}$
2. If $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2}, \ldots . . . \mathrm{Cn}$ denote the coefficients in the binomial expansion of $(1+\mathrm{x})^{\mathrm{n}}$, prove that: $\mathrm{C} _{0}+3 \mathrm{C} _{1}+5 \mathrm{C} _{2}+\ldots . .+(2 \mathrm{n}+1) \mathrm{C} _{\mathrm{n}}=(\mathrm{n}+1) \cdot 2^{\mathrm{n}}$.
Show Answer
Solution :
We have,
$\mathrm{C} _{0}+3 \mathrm{C} _{1}+5 \mathrm{C} _{2}+\ldots .+(2 \mathrm{n}+1) \mathrm{C} _{\mathrm{n}}$
$=\sum _{\mathrm{r}=0}^{\mathrm{n}}(2 \mathrm{r}+1) \mathrm{C} _{\mathrm{r}}$
$=\sum _{\mathrm{r}=0}^{\mathrm{n}}(2 \mathrm{r}+1)^{\mathrm{n}} \mathrm{C} _{\mathrm{r}} \quad\left[\because \mathrm{C} _{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}\right]$
$=\sum _{\mathrm{r}=0}^{\mathrm{n}}\left(2 \mathrm{r} \cdot{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}\right)$
$=\sum _{\mathrm{r}=0}^{\mathrm{n}} 2 \mathrm{r} \cdot \mathrm{n} _{\mathrm{r}}+\sum _{\mathrm{r}=0}^{\mathrm{n}}{ }^{n} C _{\mathrm{r}}$
$=2 \sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r} \cdot \frac{\mathrm{n}}{\mathrm{r}} \cdot{ }^{\mathrm{n}-1} \mathrm{C} _{\mathrm{r}-1}+\sum _{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}} \quad\left[\because{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}=\frac{\mathrm{n}}{\mathrm{r}} \cdot{ }^{\mathrm{n}-1} \mathrm{C} _{\mathrm{r}-1}\right]$
$ \begin{aligned} & =2 n \sum _{r=1}^{n}{ }^{n-1} C _{r-1}+\sum _{r=0}^{n} C _{r} \\ & =2 n \cdot 2^{n-1}+2^{n} \\ & =n \cdot 2^{n}+2^{n}=(n+1) 2^{n} \end{aligned} $
3. If $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2}, \ldots . . \mathrm{Cn}$ denote the coefficients in the binomial expansion of $(1+\mathrm{x})^{\mathrm{n}}$, prove that: $1^{2} \cdot \mathrm{C} _{1}+2^{2} \cdot \mathrm{C} _{2}+3^{2} \cdot \mathrm{C} _{3}+\ldots .+\mathrm{n}^{2} \cdot \mathrm{C} _{\mathrm{n}}=\mathrm{n}(\mathrm{n}+1) 2^{\mathrm{n}-2}$
Show Answer
Solution :
We have,
$1^{2} \cdot \mathrm{C} _{1}+2^{2} \cdot \mathrm{C} _{2}+3^{2} \cdot \mathrm{C} _{3}+\ldots . .+\mathrm{n}^{2} \cdot \mathrm{C} _{\mathrm{n}}$
$=\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{2} \cdot \mathrm{C} _{\mathrm{r}}$
$=\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{2} \cdot \mathrm{n} _{\mathrm{r}}$
$=\sum _{\mathrm{r}=1}^{\mathrm{n}}[\mathrm{r}(\mathrm{r}-1)+\mathrm{r}]{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}$
$=\sum _{r=1}^{n} r(r-1) \cdot \frac{n}{r} \cdot \frac{n-1}{r-1}{ }^{n-2} C _{r-2}+\sum _{r=1}^{n} r \cdot \frac{n}{r}{ }^{n-1} C _{r-1}$
$=n(n-1)\left(\sum _{r=2}^{n}{ }^{n-2} C _{r-2}\right)+n\left(\sum _{r=1}^{n} n-1 C _{r-1}\right)$
$=\mathrm{n}(\mathrm{n}-1)\left({ }^{\mathrm{n}-2} \mathrm{C} _{0}+{ }^{+\mathrm{n}-2} \mathrm{C} _{1}+\ldots .+{ }^{+\mathrm{n}-2} \mathrm{C} _{\mathrm{n}-2}\right)$
$+n\left({ }^{(n-1} \mathrm{C} _{0}{ }^{n-1} \mathrm{C} _{1}+\ldots .+{ }^{\mathrm{n}-1} \mathrm{C} _{\mathrm{n}-1}\right)$
$=\mathrm{n}(\mathrm{n}-1) \cdot 2^{\mathrm{n}-2}+\mathrm{n} \cdot 2^{\mathrm{n}-1}$
$=\mathrm{n}(\mathrm{n}-1+2) 2^{\mathrm{n}-2}$
$=\mathrm{n}(\mathrm{n}+1) 2^{\mathrm{n}-2}$
4. If $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2}, \ldots . . \mathrm{Cn}$ denote the coefficients in the binomial expansion of $(1+\mathrm{x})^{\mathrm{n}}$, prove that: $1^{3} \cdot \mathrm{C} _{1}+2^{3} \cdot \mathrm{C} _{2}+3^{3} \cdot \mathrm{C} _{3}+\ldots . .+\mathrm{n}^{3} \cdot \mathrm{C} _{\mathrm{n}}=\mathrm{n}^{2}(\mathrm{n}+3) 2^{\mathrm{n}-3}$
Show Answer
Solution :
We have,
$1^{3} \cdot \mathrm{C} _{1}+2^{3} \cdot \mathrm{C} _{2}+3^{3} \cdot \mathrm{C} _{3}+\ldots . .+\mathrm{n}^{3} \cdot \mathrm{C} _{\mathrm{n}}$
$=\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{3} \cdot \mathrm{C} _{\mathrm{r}}$
$=\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{3} \cdot{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}$
$ \begin{aligned} & =\sum _{\mathrm{r}=1}^{\mathrm{n}}[\mathrm{r}(\mathrm{r}-1)(\mathrm{r}-2)+3 \mathrm{r}(\mathrm{r}-1)+\mathrm{r}]{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}} \\ & =\sum _{r=1}^{n} r(r-1)(r-2)^{n} C _{r}+\sum _{r=1}^{n} 3 r(r-1)^{n} C _{r}+\sum _{r=1}^{n} r^{n} C _{r} \\ & =\sum _{r=3}^{n} r(r-1)(r-2) \cdot \frac{n}{r} \cdot \frac{n-1}{r-1} \cdot \frac{n-2}{r-2} \cdot{ }^{n-3} C _{r-3} \\ & +\sum _{r=2}^{n} 3 r(r-1) \frac{n}{r} \cdot \frac{n-1}{r-1}{ }^{n-2} C _{r-2}+\sum _{r=1}^{n} r \cdot \frac{n _{n}-1}{r} C _{r-1} \\ & =n(n-1)(n-2)\left(\sum _{r=3}^{n}{ }^{n-3} C _{r-3}\right)+3 n(n-1)\left(\sum _{r=2}^{n}{ }^{n-2} C _{r-2}\right)+n\left(\sum _{r=1}^{n} n-1 C _{r-1}\right) \\ & =\quad n(n-1)(n-2)\left\{{ }^{n-3} C _{0}+{ }^{n-3} C _{1}+\ldots \ldots .{ }^{n-3} C _{n-3}\right. \\ & +3 n(n-1)\left({ }^{n-2} C _{0}+{ }^{n-2} C _{1}+\ldots \ldots .{ }^{n-2} C _{n-2}\right. \\ & +n\left\{{ }^{n-1} C _{0}+{ }^{n-1} C _{1}+\ldots \ldots+{ }^{n-1} C _{n-1}\right\} \\ & =\quad n(n-1)(n-2) \cdot 2^{n-3}+3 n(n-1) \cdot 2^{n-2}+n \cdot 2^{n-1} \\ & =\{(\mathrm{n}-1)(\mathrm{n}-2)+6(\mathrm{n}-1)+4\} \mathrm{n} 2^{\mathrm{n}-3} \\ & =n\left(n^{2}+3 n\right) 2^{n-3} \\ & =\mathrm{n}^{2}(\mathrm{n}+3) 2^{\mathrm{n}-3} \end{aligned} $
5. If $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2} \ldots \ldots, \mathrm{C} _{\mathrm{n}-1}, \mathrm{C} _{\mathrm{n}}$ denote the binomial coefficients in the expansion of $(1+\mathrm{x})^{\mathrm{n}}$, prove that: $\frac{\mathrm{C} _{1}}{\mathrm{C} _{0}}+2 \cdot \frac{\mathrm{C} _{2}}{\mathrm{C} _{1}}+3 \cdot \frac{\mathrm{C} _{3}}{\mathrm{C} _{2}}+\ldots \ldots+\mathrm{n} \cdot \frac{\mathrm{C} _{\mathrm{n}}}{\mathrm{C} _{\mathrm{n}-1}}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
Show Answer
Solution : We have, $\frac{C _{1}}{C _{0}}+2 \cdot \frac{C _{2}}{C _{1}}+3 \cdot \frac{C _{3}}{C _{2}}+\ldots \ldots+n \cdot \frac{C _{n}}{C _{n-1}}$
$ \begin{aligned} & =\sum _{r=1}^{n} r \frac{C _{r}}{C _{r-1}} \\ & =\sum _{r=1}^{n} r \cdot \frac{{ }^{n} C _{r}}{{ }^{n} C _{r-1}} \\ & =\sum _{r=1}^{n} r \cdot\left(\frac{n-r+1}{r}\right) \quad\left[\because \frac{{ }^{n} C _{r}}{{ }^{n} C _{r-1}}=\frac{n-r+1}{r}\right] \\ & =\sum _{r=1}^{n}(n-r+1) \\ & =\sum _{r=1}^{n}\{(n+1)-r\} \end{aligned} $
$ \begin{aligned} & =\mathrm{n}(\mathrm{n}+1)-\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r} \\ & =\mathrm{n}(\mathrm{n}+1)-\frac{\mathrm{n}(\mathrm{n}+1)}{2} \\ & =\frac{\mathrm{n}(\mathrm{n}+1)}{2} \end{aligned} $
6. If $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2} \ldots . ., \mathrm{C} _{\mathrm{n}-1}, \mathrm{C} _{\mathrm{n}}$ denote the binomial coefficients in the expansion of $(1+\mathrm{x})^{\mathrm{n}}$, prove that: $\left(\mathrm{C} _{0}+\mathrm{C} _{1}\right)\left(\mathrm{C} _{1}+\mathrm{C} _{2}\right)\left(\mathrm{C} _{2}+\mathrm{C} _{3}\right)\left(\mathrm{C} _{3}+\mathrm{C} _{4}\right) \ldots \ldots\left(\mathrm{C} _{\mathrm{n}-1}+\mathrm{C} _{\mathrm{n}}\right)=\frac{\mathrm{C} _{0} \mathrm{C} _{1} \mathrm{C} _{2} \ldots . \mathrm{C} _{\mathrm{n}-1}(\mathrm{n}+1)^{\mathrm{n}}}{\mathrm{n} !}$
Show Answer
Solution :
We, have $\left(\mathrm{C} _{0}+\mathrm{C} _{1}\right)\left(\mathrm{C} _{1}+\mathrm{C} _{2}\right)\left(\mathrm{C} _{2}+\mathrm{C} _{3}\right) \ldots . .\left(\mathrm{C} _{\mathrm{n}-1}+\mathrm{C} _{\mathrm{n}}\right)$
$=\mathrm{C} _{0} \mathrm{C} _{1} \mathrm{C} _{2} \ldots . . \mathrm{C} _{\mathrm{n}-1}\left(1+\frac{\mathrm{C} _{1}}{\mathrm{C} _{0}}\right)\left(1+\frac{\mathrm{C} _{2}}{\mathrm{C} _{1}}\right) \cdots\left(1+\frac{\mathrm{C} _{\mathrm{n}}}{\mathrm{C} _{\mathrm{n}-1}}\right)$
$=\left(C _{0} C _{1} \ldots . C _{n-1}\right) \prod _{r=1}^{n}\left\{1+\frac{{ }^{n} C _{r}}{{ }^{n} C _{r-1}}\right\}$
$=\left(\mathrm{C} _{0} \mathrm{C} _{1} \ldots \mathrm{C} _{\mathrm{n}-1}\right) \prod _{\mathrm{r}=1}^{\mathrm{n}}\left\{1+\frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}\right\}$
$=\left(\mathrm{C} _{0} \mathrm{C} _{1} \ldots \mathrm{C} _{\mathrm{n}-1}\right) \prod _{\mathrm{r}=1}^{\mathrm{n}}\left(\frac{\mathrm{n}+1}{\mathrm{r}}\right)$
$=\left(\mathrm{C} _{0} \mathrm{C} _{1} \ldots \mathrm{C} _{\mathrm{n}-1}\right) \prod _{\mathrm{r}=1}^{\mathrm{n}} \frac{(\mathrm{n}+1)^{\mathrm{n}}}{\mathrm{n} !}$
7. If $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2}, \ldots . . \mathrm{Cn}$ denote the binomial cofficients in the expansion of $(1+\mathrm{x})^{\mathrm{n}}$, prove that $\mathrm{C} _{0}+\frac{\mathrm{C} _{1}}{2}+\frac{\mathrm{C} _{2}}{3}+\ldots \ldots+\frac{\mathrm{C} _{\mathrm{n}}}{\mathrm{n}+1}=\frac{2^{\mathrm{n}+1}-1}{\mathrm{n}+1}$
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Solution : we have, $\mathrm{C} _{0}+\frac{\mathrm{C} _{1}}{2}+\frac{\mathrm{C} _{2}}{3}+\ldots \ldots .+\frac{\mathrm{C} _{\mathrm{n}}}{\mathrm{n}+1}$
$ \begin{aligned} & =\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{C} _{\mathrm{r}}}{\mathrm{r}+1} \\ & =\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{1}{\mathrm{r}+1} \cdot{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}} \\ & =\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{1}{\mathrm{n}+1} \cdot \frac{\mathrm{n}+1}{\mathrm{r}+1} \cdot{ }^{n} \mathrm{C} _{\mathrm{r}} \end{aligned} $
$\begin{aligned} & =\frac{1}{n+1} \sum_{r=0}^n \frac{n+1}{r+1} \cdot{ }^n C_r \\ & =\frac{1}{n+1} \sum_{r=0}^n{ }^{n+1} C_{r+1} \qquad\left[{}^{n+1}C_{r+1}=\frac{n+1}{r+1}.{}^n C_r\right]\\ & =\frac{1}{n+1}\left[{ }^{n+1} C_1+{ }^{n+1} C_2+{ }^{n+1} C_3+\ldots .{ }^{n+1} C_{n+1}\right] \\ & =\frac{1}{n+1}\left[{ }^{n+1} C_r+{ }^{n+1} C_1+\ldots .{ }^{n+1} C_{n+1}-\left({ }^{n+1} C_0\right)\right] \\ & =\frac{1}{n+1}\left[2{ }^{n+1}-1\right]\end{aligned}$
Practice questions
1. The sum $\frac{1}{1 !(\mathrm{n}-1) !}+\frac{1}{3 !(\mathrm{n}-3) !}+\frac{1}{5 !(\mathrm{n}-5) !}+\ldots \ldots \ldots=$
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Answer: $\frac{2^{n-1}}{n!}$2. If the coefficient of $x^{n}$ in $(1+x)^{101}\left(1-x+x^{2}\right)^{100}$ is non-zero, then $n$ cannot be of the form
(a). $3 \mathrm{r}+1$
(b). $3 \mathrm{r}$
(c). $3 \mathrm{r}+2$
(d). none of these
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Answer: (c)3. The coefficient of $x^{r} ; 0 \leq r \leq n-1$, is the expansion of $(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-2}$ $(\mathrm{x}+2)^{2}+\ldots . .(\mathrm{x}+2)^{\mathrm{n}-1}$ are
(a). ${ }^{n} C _{r}\left(3^{r}-2^{n}\right)$
(b). ${ }^{n} C _{r}\left(3^{n-r}-2^{n-r}\right)$
(c). ${ }^{n} C _{r}\left(3^{r}-2^{n-r}\right)$
(d). none of these
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Answer: (b)4. The number of real negative terms in the binomial expension of $(1+i x)^{4 n-2}, n \in N, x>0$ is
(a). $\mathrm{n}$
(b). $n+1$
(c). $\mathrm{n}-1$
(d). $2 \mathrm{n}$
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Answer: (a)5. $(\mathrm{n}+2){ }^{n} C _{0} 2^{\mathrm{n}+1}-(\mathrm{n}+1){ }^{\mathrm{n}} \mathrm{C} _{1} 2^{\mathrm{n}}+\mathrm{n} \cdot{ }^{n} \mathrm{C} _{2} 2^{\mathrm{n}-1} \ldots . .$. is equal to
(a). $4$
(b). $4 \mathrm{n}$
(c). $4(\mathrm{n}+1)$
(d). $2(\mathrm{n}+2)$
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Answer: (c)6. $\sum _{\mathrm{k}=1}^{\infty} \mathrm{k}\left(1+\frac{1}{\mathrm{n}}\right)^{\mathrm{k}-1}=$
(a). $\mathrm{n}(\mathrm{n}-1)$
(b). $\mathrm{n}(\mathrm{n}+1)$
(c). $n^{2}$
(d). $(\mathrm{n}+1)^{2}$
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Answer: (c)7. The sum of rational term in $(\sqrt{2}+\sqrt[3]{3}+\sqrt[6]{5})^{10}$ is equal to
(a). 12632
(b). 1260
(c). 126
(d). none of these
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Answer: (d)8. Last two digit of $(23)^{14}$ are
(a). 01
(b). 03
(c). 09
(d). none of these
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Answer: (c)9. If $(4+\sqrt{15})^{\mathrm{n}}=\mathrm{I}+\mathrm{f}$, where $\mathrm{n}$ is an odd natural number, $\mathrm{I}$ is an interger and $0<\mathrm{f}<1$, then
(a). I is a natural number
(b). I an even integer
(c). $(\mathrm{I}+\mathrm{f})(1-\mathrm{F})=1$
(d). $ 1-\mathrm{f}=(4+\sqrt{5})^{\mathrm{n}}$
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Answer: (a, c, d)10. The number of rational numbers lying in the interval $(2002, 2003)$ all whose digits after the decimal point are non-zero and are in decreasing order is
(a). $ \sum _{\mathrm{i}=1}^{9}{ }^{9} \mathrm{P} _{\mathrm{i}}$
(b). $ \sum _{\mathrm{i}=1}^{10}{ }^{9} \mathrm{P} _{\mathrm{i}}$
(c). $ 2^{9}-1$
(d). $2^{10}-1$
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Answer: (c)11. Match the following:
Column I | Column II |
---|---|
(a). ${ }^{\mathrm{m}} \mathrm{C} _{1}{ }^{n} \mathrm{C} _{\mathrm{m}}-{ }^{\mathrm{m}} \mathrm{C} _{2}{ }^{2 \mathrm{n}} \mathrm{C} _{\mathrm{m}}+{ }^{\mathrm{m}} \mathrm{C} _{3}{ }^{3 \mathrm{n}} \mathrm{C} _{\mathrm{m}} \cdots .$. $+(-1)^{\mathrm{m}-1 \mathrm{~m}} \mathrm{C} _{\mathrm{m}}{ }^{\mathrm{m}} \mathrm{C} _{\mathrm{m}}$ is | p. The coefficent of $\mathrm{x}^{\mathrm{m}}$ in the expansion of $\left((1+\mathrm{x})^{\mathrm{n}}-1\right)^{\mathrm{m}}$ |
(b). ${ }^{n} C _{m}+{ }^{n-1} C _{m}+{ }^{n-2} C _{m}+\ldots . .+{ }^{m} C _{m}$ is | q. The coefficent of $x^{m}$ in $\frac{(1+x)^{n+1}}{x}$ |
(c). $\mathrm{C} _{0} \mathrm{C} _{\mathrm{n}}+\mathrm{C} _{1} \mathrm{C} _{\mathrm{n}-1}+\ldots \ldots . \mathrm{C} _{\mathrm{n}} \mathrm{C} _{0}$ is | r. The coefficent of $x^{n}$ in $(1+x)^{2 n}$ |
(d). $ 2 k^{n} C _{0}-2^{k-1}{ }^{n} C _{1}{ }^{n-1} C _{k-1}+(-1)^{k}{ }^{n} C _{k}$ ${ }^{\mathrm{n}-\mathrm{k}} \mathrm{C} _{0}$ is | s. The coefficent of $x^{k}$ in the expansion $(1+\mathrm{x})^{\mathrm{n}}$ |