BINOMIAL THEOREM - 3 (Principle and simple applications - Problem Solving)
1. Binomial Theorem for positive integral index.
If $x, y \varepsilon R$ and $n \varepsilon N$
$ (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 \ldots+{ }^{n} C _{r} x^{n-r} y^{r}+\ldots .+{ }^{n} C _{n} y^{n}=\sum _{r=0}^{n}{ }^{n} C _{r} x^{n-r} y^{r} $
Properties
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Number of terms of the above expansion is $(\mathrm{n}+1)$
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The binomial coefficients equidistant from the beginning and the end in a binomial expansion are equal.
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General term $=\mathrm{T} _{\mathrm{r}+1}={ }^{n} \mathrm{C} _{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}} \mathrm{y}^{\mathrm{r}}$
Middle term $-\left[\begin{array}{l}\mathrm{n} \text { is even : only one middle term }\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }} \text { term. } \\ \mathrm{n} \text { is odd : Two middle terms }\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }} \text { and }\left(\frac{\mathrm{n}+3}{2}\right)^{\text {th }} \text { term. }\end{array}\right.$
Greatest coefficient $-\left[\begin{array}{l}n \text { is even }{ }^{n} C _{\frac{n}{2}} \\ n \text { is odd }{ }^{n} C _{\frac{n-1}{2}} \text { and }{ }^{n} C _{\frac{n+1}{2}}\end{array}\right.$
Greatest Term
To find numerically greatest term in the expansion of $(1+x)^{n}$
i. Calculate $\mathrm{m}=\frac{|\mathrm{x}|(\mathrm{n}+1)}{|\mathrm{x}|+1}$
ii. If $\mathrm{m}$ is an integer, then $\mathrm{T} _{\mathrm{m}}$ and $\mathrm{T} _{\mathrm{m}+1}$ are equal and both are greatest term.
iii. If $\mathrm{m}$ is not an integer then $\mathrm{T} _{[\mathrm{m}]+1}$ is the greatest term.
Note : To find the greatest term in the expansion of $(x+y)^{n}$, find the greatest term in $\left(1+\frac{y}{x}\right)^{n}$ and then multiply by $x^{n}\left(\operatorname{since}(x+y)^{n}=x^{n}\left(1+\frac{y}{x}\right)^{n}\right.$
2. Multinomial Theorem (for a positive integral index)
$ \left(\mathrm{x} _{1}+\mathrm{x} _{2}+\mathrm{x} _{3}+\ldots . \mathrm{x} _{\mathrm{k}}\right)^{\mathrm{n}}=\sum \frac{\mathrm{n} !}{\mathrm{n} _{1} ! \mathrm{n} _{2} ! \mathrm{n} _{3} ! \ldots . \mathrm{n} _{\mathrm{k}} !} \mathrm{x} _{1}^{\mathrm{n} _{1}} \mathrm{x} _{2}^{\mathrm{n} _{2}} \mathrm{x} _{3}{ }^{\mathrm{n} _{3}} \ldots \ldots \mathrm{x} _{\mathrm{k}}{ }^{\mathrm{n} _{k}} $
where $\mathrm{n} _{\mathrm{i}} \varepsilon \quad\{0,1,2, \ldots . \mathrm{n}\}, \mathrm{n} _{1}+\mathrm{n} _{2}+\ldots . .+\mathrm{n} _{\mathrm{k}}=\mathrm{n}$
- The greatest coefficient in this expansion is $\frac{\mathrm{n} !}{(\mathrm{q} !)^{\mathrm{kr}}((\mathrm{q}+1) !)^{\mathrm{r}}}$ where $\mathrm{q}$ is the quotient and $r$ is the remainder when $n$ is divided by $\mathrm{k}$.
$\quad$ Eg. Find the greatest coefficient in $(x+y+z+w)^{15}$
$\quad$ $\mathrm{n}=15, \mathrm{k}=4$ 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 $\mathrm{x}^{\mathrm{n}-1} \mathrm{C} _{\mathrm{k}-1}$.
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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 $x _{1}=x _{2}=\ldots . . x _{k}=1$.
3. Binomial Theorem for Negative or Fractional Indices
If $n \varepsilon Q$, then
$(1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !} \mathrm{x}^{2}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !} \mathrm{x}^{2}+\ldots . . \infty$ provided $|\mathrm{x}|<1$.
- For any index $n$, the general term in the expansion of
$\quad$$\quad$ i. $\quad(1+x)^{n}$ is $\mathrm{T} _{\mathrm{r}+1}=\frac{\mathrm{n}(\mathrm{n}-1) \ldots \ldots \ldots .(\mathrm{n}-\mathrm{r}+1)}{\mathrm{r} !} \mathrm{x}^{\mathrm{r}}$
$\quad$$\quad$ ii. $\quad(1+\mathrm{x})^{-\mathrm{n}}$ is $\mathrm{T} _{\mathrm{r}+1}=\frac{(-1)^{\mathrm{r}} \mathrm{n}(\mathrm{n}+1) \ldots \ldots \ldots .(\mathrm{n}+\mathrm{r}-1)}{\mathrm{r} !} \mathrm{x}^{\mathrm{r}}$
$\quad$$\quad$ iii. $\quad(1-x)^{n}$ is $T _{r+1}=\frac{(-1)^{r} n(n-1) \ldots \ldots \ldots .(n-r+1)}{r !} x^{r}$
$\quad$$\quad$ iv. $\quad(1-\mathrm{x})^{-n}$ is $\mathrm{T} _{\mathrm{r}+1}=\frac{\mathrm{n}(\mathrm{n}+1) \ldots \ldots \ldots .(\mathrm{n}+\mathrm{r}-1)}{\mathrm{r} !} \mathrm{x}^{\mathrm{r}}$
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The following expansions should be remembered (for $|\mathrm{x}|<1$ ). i. $\quad(1+x)^{-1}=1-x+x^{2}-x^{3}+$ $\infty$ ii. $\quad(1-x)^{-1}=1+x+x^{2}+x^{3}+$ $\infty$ iii. $\quad(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+$ $\infty$ iv. $\quad(1-x)^{-2}=1+2 x+3 x^{2}+4 x^{3}+$ $\infty$
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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 $\frac{1}{\mathrm{x}}$, which then will be small.
4. Exponential series
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$e^{x}=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots . . \infty$
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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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$\mathrm{e}+\mathrm{e}^{-1}=2\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots ..\right)$
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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)$
5. 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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$\log _{e}(1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots . \infty,-1 \leq x<1$ or $|x|<1$
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$\log \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots . \infty\right),-1<x<1$
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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. Let $(1+\mathrm{x})^{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}} \mathrm{a} _{\mathrm{r}} \mathrm{x}^{\mathrm{r}}$,
then $\left(1+\frac{a _{1}}{a _{0}}\right)\left(1+\frac{a _{2}}{a _{1}}\right)\left(1+\frac{a _{3}}{a _{2}}\right) \ldots \ldots .\left(1+\frac{a _{n}}{a _{n-1}}\right)$ is equal to
(a). $\frac{(\mathrm{n}+1)^{\mathrm{n}+1}}{\mathrm{n} !}$
(b). $\frac{(\mathrm{n}+1)^{\mathrm{n}}}{\mathrm{n} !}$
(c). $\frac{(n)^{n-1}}{(n-1) !}$
(d). None of these
Show Answer
Solution :
$ \begin{aligned} (1+x)^{n} & ={ }^{n} C _{0}+{ }^{n} C _{1} x+{ }^{n} C _{2} x^{2}+\ldots \ldots \ldots .+{ }^{n} C _{n} x^{n} \\ & =a _{0}+a _{1} x+a _{2} x^{2}+\ldots \ldots \ldots . .+a _{n} x^{n}(\text { given }) \end{aligned} $
Comparing $\mathrm{a} _{0}={ }^{n} \mathrm{C} _{0}, \mathrm{a} _{1}={ }^{\mathrm{n}} \mathrm{C} _{1}, \mathrm{a} _{2}={ }^{\mathrm{n}} \mathrm{C} _{2}, \ldots \ldots \ldots .$. and so on.
$ \begin{aligned} & \therefore\left(1+\frac{{ }^{n} C _{1}}{{ }^{n} C _{0}}\right)\left(1+\frac{{ }^{n} C _{2}}{{ }^{n} C _{1}}\right)\left(1+\frac{{ }^{n} C _{3}}{{ }^{n} C _{2}}\right) \ldots \ldots .\left(1+\frac{{ }^{n} C _{n}}{{ }^{n} C _{n-1}}\right) \\ & =\left(1+\frac{n-0}{1}\right)\left(1+\frac{n-1}{2}\right)\left(1+\frac{n-2}{3}\right) \ldots \ldots .\left(1+\frac{n-n+1}{n}\right) \\ & =\frac{1+n}{1} \cdot \frac{1+n}{2} \cdot \frac{1+n}{3} \ldots \ldots . \frac{1+n}{n}=\frac{(n+1)^{n}}{n !} \end{aligned} $
Answer: b
2. If $\mathrm{a} _{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{1}{{ }^{n} \mathrm{C} _{\mathrm{r}}}$, then $\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{r}}{{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}}$ equals
(a). $(n-1) \cdot a _{n}$
(b). $ n a _{n}$
(c). $\frac{1}{2} \mathrm{na} _{\mathrm{n}}$
(d). None of these
Show Answer
Solution :
Let $S=\sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{r}}{{ }^{n} \mathrm{C} _{\mathrm{r}}}$
$\Rightarrow S=\frac{0}{{ }^{n} C _{0}}+\frac{1}{{ }^{n} C _{1}}+\frac{2}{{ }^{n} C _{2}}+\frac{3}{{ }^{n} C _{3}}+\ldots \ldots \ldots+\frac{n}{{ }^{n} C _{n}}…….(1)$
Also, $S=\frac{n}{{ }^{n} C _{0}}+\frac{n-1}{{ }^{n} C _{1}}+\frac{n-2}{{ }^{n} C _{2}}+\frac{n-3}{{ }^{n} C _{3}}+\ldots \ldots \ldots+\frac{0}{{ }^{n} C _{n}}……..(2)$
Adding (1) and 2
$ \begin{aligned} & 2 S=\frac{n}{{ }^{n} C _{0}}+\frac{n}{{ }^{n} C _{1}}+\frac{n}{{ }^{n} C _{2}}+\ldots \ldots .+\frac{n}{{ }^{n} C _{n}} \\ & 2 S=n\left(\frac{1}{{ }^{n} C _{0}}+\frac{1}{{ }^{n} C _{1}}+\frac{1}{{ }^{n} C _{2}}+\ldots \ldots \ldots+\frac{1}{{ }^{n} C _{n}}\right) \\ & \Rightarrow 2 S=n a _{n} \Rightarrow S=\frac{1}{2} n a _{n} \end{aligned} $
Answer: c
3. If $(1+x)^{10}=a _{0}+a _{1} x+a _{2} x^{2}+a _{3} x^{3}+\ldots \ldots .+a _{10} x^{10}$, then $\left(a _{0}-a _{2}+a _{4}-a _{6}+a _{8}-a _{10}\right)^{2}+\left(a _{1}-a _{3}+a _{5}-a _{7}+a _{9}\right)^{2}$ is equal to
(a). $3^{10}$
(b). $2^{10}$
(c). $2^{9}$
(d). none of these
Show Answer
Solution:
Put $\mathrm{x}=\mathrm{i}$ and $\mathrm{x}=-\mathrm{i}$
$\Rightarrow(1+i)^{10}=\left(a _{0}-a _{2}+a _{4}-a _{6}+a _{8}-a _{10}\right)+i\left(a _{1}-a _{3}+a _{5}+a _{5}-a _{7}+a _{9}\right)……(1)$
Also, $(1-i)^{10}=\left(a _{0}-a _{2}+a _{4}-a _{6}+a _{8}-a _{10}\right)-i\left(a _{1}-a _{3}+a _{5}-a _{7}+a _{9}\right)…….(2)$
Multiply (1) and (2)
$\left((1+i)(1-i)^{10}=\left(a _{0}-a _{2}+a _{4}-a _{6}+a _{8}-a _{10}\right)^{2}+\left(a _{1}-a _{3}+a _{5}-a _{7}+a _{9}\right)^{2}\right.$
$2^{10}=\left(a _{0}-a _{2}+a _{4}-a _{6}+a _{8}-a _{10}\right)^{2}+\left(a _{1}-a _{3}+a _{5}-a _{7}+a _{9}\right)^{2}$
Answer: b
4. If $\left(1+x+2 x^{2}\right)^{20}=a _{0}+a _{1} x+a _{2} x^{2}+a _{3} x^{3}+\ldots \ldots .+a _{40} x^{40}$, then $a _{0}+a _{2}+a _{4}+\ldots \ldots .+a _{38}$ is equal to
(a). $2^{19}\left(2^{20}-1\right)$
(b). $ 2^{20}\left(2^{19}-1\right)$
(c). $ 2^{19}\left(2^{20}+1\right)$
(d). none of these
Show Answer
Solution :
Put $\mathrm{x}=1$ and $\mathrm{x}=-1$ and adding we get $4^{20}+2^{20}=2\left(\mathrm{a} _{0}+\mathrm{a} _{2}+\mathrm{a} _{4}+\ldots+\mathrm{a} _{38}+\mathrm{a} _{40}\right)$
$\Rightarrow 2^{39}+2^{19}=\mathrm{a} _{0}+\mathrm{a} _{2}+\ldots .+\mathrm{a} _{38}+2^{20} \quad\left(\because \mathrm{a} _{40}=2^{20}\right)$
$\therefore \mathrm{a} _{0}+\mathrm{a} _{2}+\mathrm{a} _{4}+\ldots . .+\mathrm{a} _{38}=2^{39}+2^{19}-2^{20}$
$=2^{19}\left(2^{20}+1-2\right)$
$=2^{19}\left(2^{20}-1\right)$
Answer: a
5. The coefficient of $x^{13}$ in the expansion of $(1-x)^{5}\left(1+x+x^{2}+x^{3}\right)^{4}$ is
(a). $4$
(b). $-4$
(c). $0$
(d). none of these
Show Answer
Solution :
Coefficient of $x^{13}$ in $=(1-x)^{5}\left(1+x+x^{2}+x^{3}\right)^{4}=(1-x)^{5}\left((1+x)\left(1+x^{2}\right)\right)^{4}$
$=(1-\mathrm{x})\left\{(1-\mathrm{x})(1+\mathrm{x})\left(1+\mathrm{x}^{2}\right)\right\}^{4}$
$=(1-\mathrm{x})\left\{\left(1-\mathrm{x}^{4}\right)^{4}\right.$
$=(1-\mathrm{x})\left(1-4 \mathrm{x}^{4}+6 \mathrm{x}^{8}-4 \mathrm{x}^{12}+\mathrm{x}^{16}\right)$
$\therefore \quad$ coefficient of $\mathrm{x}^{13}$ is $-1 \times-4=4$
Answer: a
6. The sum ${ }^{20} \mathrm{C} _{0}+{ }^{20} \mathrm{C} _{1}+{ }^{20} \mathrm{C} _{2}+\ldots \ldots \ldots \ldots . .+{ }^{20} \mathrm{C} _{10}$ is equal to
(a). $2^{20}+\frac{20 !}{(10 !)^{2}}$
(b). $2^{19}-\frac{1}{2} \frac{20 !}{(10 !)^{2}}$
(c). $2^{19}+\frac{20 !}{(10 !)^{2}}$
(d). none of these
Show Answer
Solution :
In the expansion of $(1+x)^{20}$, put $x=1$
$2^{20}={ }^{20} \mathrm{C} _{0}+{ }^{20} \mathrm{C} _{1}+{ }^{20} \mathrm{C} _{2}+\ldots \ldots \ldots \ldots . .+{ }^{20} \mathrm{C} _{9}+{ }^{20} \mathrm{C} _{10}+\ldots \ldots .{ }^{20} \mathrm{C} _{20}$
$=2\left({ }^{20} \mathrm{C} _{0}+{ }^{20} \mathrm{C} _{1}+\ldots \ldots \ldots \ldots . .+{ }^{20} \mathrm{C} _{10}\right)-{ }^{20} \mathrm{C} _{10} \quad\left(\because{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}-\mathrm{r}}\right)$
$\Rightarrow \frac{2^{20}+{ }^{20} \mathrm{C} _{10}}{2}={ }^{20} \mathrm{C} _{0}+{ }^{20} \mathrm{C} _{1}+$ $+{ }^{20} \mathrm{C} _{10}$
$=2^{19}+\frac{1}{2} \frac{20 !}{(10 !)^{2}}={ }^{20} \mathrm{C} _{0}+{ }^{20} \mathrm{C} _{1}+$. $+{ }^{20} \mathrm{C} _{10}$
Answer: d
Practice questions
1. The coefficient of $\mathrm{x}^{4}$ in $\left(\frac{\mathrm{x}}{2}-\frac{3}{\mathrm{x}^{2}}\right)^{10}$ is
(a). $\frac{405}{256}$
(b). $\frac{405}{259}$
(c). $\frac{450}{263}$
(d). None of these
Show Answer
Answer: (a)2. Let $T _{n}$ denotes the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T _{n+1}-T _{n}=21$, then $n$ equals
(a). 5
(b). 7
(c). 6
(d). 4
Show Answer
Answer: (b)3. The sum $\sum _{i=1}^{m}\left(\begin{array}{c}10 \ i\end{array}\right)\left(\begin{array}{c}20 \ m-i\end{array}\right)$, where $\left(\begin{array}{l}p \ q\end{array}\right)=0$ if $p>q$, is maximum when $m$ is
(a). 5
(b). 10
(c). 15
(d). 20
Show Answer
Answer: (c)4. Coefficient of $\mathrm{t}^{24}$ in $\left(1+\mathrm{t}^{2}\right)^{12}\left(1+\mathrm{t}^{12}\right)\left(1+\mathrm{t}^{24}\right)$ is
(a). ${}^{12} \mathrm{C} _{6}+3$
(b). ${}^{12} \mathrm{C} _{6}+1$
(c). ${}^{12} \mathrm{C} _{6}$
(d). ${}^{12} \mathrm{C} _{6}+2$
Show Answer
Answer: (d)5. If ${ }^{n-1} C _{r}=\left(k^{2}-3\right)^{n} C _{r+1}$, then $k$ belongs to
(a). $(-\infty,-2]$
(b). $[2, \infty)$
(c). $[-\sqrt{3}, \sqrt{3}]$
(d). $(\sqrt{3}, 2)$
Show Answer
Answer: (d)6. If $(1+a x)^{n}=1+8 x+24 x^{2}+……..,$ then $\mathrm{a}=……..$ and $\mathrm{n}=………$
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Answer: $a=2, n=4$7. The greatest term in the expansion of $\sqrt{3}\left(1+\frac{1}{\sqrt{3}}\right)^{20}$ is
(a). $\left(\begin{array}{c}20 \ 7\end{array}\right) \frac{1}{27}$
(b). $\left(\begin{array}{c}20 \ 6\end{array}\right) \frac{1}{81}$
(c). $\frac{1}{9}\left(\begin{array}{c}20 \ 9\end{array}\right)$
(d). none of these
Show Answer
Answer: (a)8. If $x=(\sqrt{2}+1)^{6}$, then the integral part of $[x]$ is
(a). 98
(b). 197
(c). 196
(d). 198
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Answer: (b)9. The greatest integer $m$ such that $5^{m}$ divides $7^{2 n}+2^{3 n-3} \cdot 3^{n-1}$ for $n \varepsilon N$ is
(a). 0
(b). 1
(c). 2
(d). 3
Show Answer
Answer: (c)10. If $\frac{1}{(1-a x)(1-b x)}=a _{0}+a _{1}+a _{2} x^{2} \ldots .$. , then $a _{n}=$
(a). $\frac{a^{n+1}-b^{n+1}}{b-a}$
(b). $\frac{b^{n+1}-a^{n+1}}{b-a}$
(c). $\frac{b^{n}-a^{n}}{b-a}$
(d). $\frac{a^{n}-b^{n}}{b-a}$
Show Answer
Answer: (b)11. Read the passage and answer the following questions.
If $\mathrm{n}$ is a positive integer and $\mathrm{a} _{1}, \mathrm{a} _{2}, \mathrm{a} _{3} \ldots . \mathrm{a} _{\mathrm{m}} \varepsilon \mathrm{C}$, then $\left(a _{1}+a _{2}+a _{3}+\ldots . .+a _{m}\right)^{n}=\sum \frac{n !}{n _{1} ! n _{2} ! n _{3} ! \ldots \ldots . n _{m} !} \cdot a _{1}{ }^{n _{1}} \cdot a _{2}{ }^{n _{2}} \cdot a _{3}{ }^{n _{3}} \ldots \ldots . a _{m}{ }^{n _{m}}$ where $n _{1}, n _{2}, n _{3} \ldots . . n _{m}$ are all non-negative integers subject to the condition $\mathrm{n} _{1}+\mathrm{n} _{2}+\mathrm{n} _{3}+\ldots .+\mathrm{n} _{\mathrm{m}}=\mathrm{n}$.
i. The number of distinct terms in the expansion of $\left(\mathrm{x} _{1}+\mathrm{x} _{2}+\mathrm{x} _{3}+\ldots .+\mathrm{x} _{\mathrm{n}}\right)^{4}$ is
(a). ${ }^{n+1} C _{4}$
(b). ${ }^{n+2} C _{4}$
(c). ${ }^{n+3} C _{4}$
(d). ${ }^{\mathrm{n}+4} \mathrm{C} _{4}$
Show Answer
Answer: (c)ii. The coefficient of $x^{3} y^{4} z$ in the expansion of $(1+x-y+z)^{9}$ is
(a). 2320
(b). 2420
(c). 2520
(d). 2620
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Answer: (c)iii. The coefficient of $a^{3} b^{4} c^{5}$ in the expansion of $(b c+c a+a b)^{6}$ is
(a). 40
(b). 60
(c). 80
(d). 100
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Answer: (b)iv. The coefficient of $x^{39}$ in the expansion of $\left(1+x+2 x^{2}\right)^{20}$ is
(a). $5.2^{19}$
(b). $ 5.2^{20}$
(c). $ 5.2^{21}$
(d). $5.2^{23}$
Show Answer
Answer: (c)v. The coefficient of $x^{20}$ in $\left(1-x+x^{2}\right)^{20}$ and in $\left(1+x-x^{2}\right)^{20}$ are respectively $a$ and $b$, then
(a). $ \mathrm{a}=\mathrm{b}$
(b). $ \mathrm{a}>\mathrm{b}$
(c). $ \mathrm{a}<\mathrm{b}$
(d). $a+b=0$
Show Answer
Answer: (b)12. Match the following:
| Column I | Column II |
|---|---|
| (a). If $\mathrm{n}$ be the degree of the polynomial $\sqrt{\left(3 x^{2}+1\right)}\left(\left(x+\sqrt{\left(3 x^{2}+1\right)}\right)^{7}-\left(x-\sqrt{\left(3 x^{2}+1\right)}\right)^{7}\right)$then $\mathrm{n}$ is divisible by | (p). 2 |
| (b). In the expression of $(x+a)^{n}$ there is only one middle term for$\mathrm{x}=3, \mathrm{a}=2$ and seventh term is numerically greatest term, then $\mathrm{n}$ is divisible by | (q). 4 |
| (c). The sum of the binomial coefficients in the expansion of $\left(\mathrm{x}^{-3 / 4}+\mathrm{nx}^{5 / 4}\right)^{\mathrm{m}}$,where $\mathrm{m}$ is positive integer lies between 200 and 400 and the term independent of$x$ is equals 448 . Then $\mathrm{n}^{5}$ is divisible by | (r). 8 |
| (s). 16 | |
| (t). 32 |
