Binomial Theorem - Principle and simple applications (Lecture-01)
1. Binomial Theorem for positive integral index.
If $x, y \varepsilon R$ and $n \varepsilon N$
$(x+y)^{n}={ }^{n} C _{0} x^{n}+{ }^{n} C _{1} x^{n-1} y+{ }^{n} C _{2} x^{n-2} y^{2}+\ldots \ldots \ldots+{ }^{n} C _{r} x^{n-r} y^{r}+\ldots .+{ }^{n} C _{n} y^{n}=\sum\limits _{r=0}^{n}{ }^{n} C _{r} x^{n-r} y^{r}$
Properties
- Number of terms of the above expansion is $(\mathrm{n}+1)$
- The binomial coefficients equidistant from the beginning and the end in a binomial expansion are equal.
- General term $=\mathrm{T} _{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}} \mathrm{y}^{\mathrm{r}}$
Middle term $-\left[\begin{array}{l}\mathrm{n} \text { is even : only one middle term }\left(\dfrac{\mathrm{n}}{2}+1\right)^{\text {th }} \text { term. } \\ \mathrm{n} \text { is odd : Two middle terms }\left(\dfrac{\mathrm{n}+1}{2}\right)^{\text {th }} \text { and }\left(\dfrac{\mathrm{n}+3}{2}\right)^{\text {th }} \text { term. }\end{array}\right.$
Greatest coefficient $-\left[\begin{array}{l}n \text { is even }{ }^{n} C _{\dfrac{n}{2}} \\ n \text { is odd }{ }^{n} C _{\dfrac{n-1}{2}} \text { and }{ }^{n} C _{\dfrac{n+1}{2}}\end{array}\right.$
Greatest Term
To find numerically greatest term in the expansion of $(1+x)^{n}$
i. Calculate $\mathrm{m}=\dfrac{|\mathrm{x}|(\mathrm{n}+1)}{|\mathrm{x}|+1}$
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 $\left(1+\dfrac{y}{x}\right)^{n}$ and then multiply by $x^{n}\left(\operatorname{since}(x+y)^{n}=x^{n}\left(1+\dfrac{y}{x}\right)^{n}\right.$
2. Multinomial Theorem (for a positive integral index)
$\left(\mathrm{x} _{1}+\mathrm{x} _{2}+\mathrm{x} _{3}+\ldots . \mathrm{x} _{\mathrm{k}}\right)^{\mathrm{n}}=\sum \dfrac{\mathrm{n} !}{\mathrm{n} _{1} ! \mathrm{n} _{2} ! \mathrm{n} _{3} ! \ldots . \mathrm{n} _{\mathrm{k}} !} \mathrm{x} _{1}^{\mathrm{n} _{1}} \mathrm{x} _{2}^{\mathrm{n} _{2}} \mathrm{x} _{3}{ }^{\mathrm{n} _{3}} \ldots . . \mathrm{x} _{\mathrm{k}}{ }^{\mathrm{n} _{k}}$
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 $\dfrac{\mathrm{n} !}{(\mathrm{q} !)^{\mathrm{k}-\mathrm{r}}((\mathrm{q}+1) !)^{\mathrm{r}}}$ where $\mathrm{q}$ is the quotient and $r$ is the remainder when $n$ is divided by $\mathrm{k}$.
Eg. Find the greatest coefficient in $(x+y+z+w)^{15}$
$\mathrm{n}=15, \mathrm{k}=4$ we have $15=4 \times 3+3$ i.e. $\mathrm{q}=3, \mathrm{r}=3$ greatest coefficient $=\dfrac{15 !}{(3 !)^{1}(4 !)^{3}}$
- Number of distinct terms in the expansion is ${ }^{n+k-1} C _{k-1}$ (Total number of terms).
- Number of positive integer solutions of $\mathrm{x} _{1}+\mathrm{x} _{2}+\ldots .+\mathrm{x} _{\mathrm{k}}=n$ is $\mathrm{x}^{\mathrm{n}-1} \mathrm{C} _{\mathrm{k}-1}$.
- Number of non negative integer solutions of $x _{1}+x _{2}+\ldots .+x _{k}=n$ is ${ }^{n+k-1} C _{k-1}$.
- 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
$(1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}+\dfrac{\mathrm{n}(\mathrm{n}-1)}{2 !} \mathrm{x}^{2}+\dfrac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !} \mathrm{x}^{2}+\ldots \ldots \infty$ provided $|\mathrm{x}|<1$.
- For any index $n$, the general term in the expansion of
i. $(1+x)^{n}$ is $T _{r+1}=\dfrac{n(n-1) \ldots \ldots \ldots .(n-r+1)}{r !} x^{r}$
ii. $(1+\mathrm{x})^{-\mathrm{n}}$ is $\mathrm{T} _{\mathrm{r}+1}=\dfrac{(-1)^{\mathrm{r}} \mathrm{n}(\mathrm{n}+1) \ldots \ldots \ldots .(\mathrm{n}+\mathrm{r}-1)}{\mathrm{r} !} \mathrm{x}^{\mathrm{r}}$
iii. $(1-x)^{n}$ is $T _{r+1}=\dfrac{(-1)^{r} n(n-1) \ldots \ldots \ldots .(n-r+1)}{r !} x^{r}$
iv. $(1-\mathrm{x})^{-\mathrm{n}}$ is $\mathrm{T} _{\mathrm{r}+1}=\dfrac{\mathrm{n}(\mathrm{n}+1) \ldots \ldots \ldots .(\mathrm{n}+\mathrm{r}-1)}{\mathrm{r} !} \mathrm{x}^{\mathrm{r}}$
- The following expansions should be remembered (for $|\mathrm{x}|<1$ ).
i. $(1+x)^{-1}=1-x+x^{2}-x^{3}+\ldots \ldots \ldots \infty$
ii. $(1-x)^{-1}=1+x+x^{2}+x^{3}+\ldots \ldots . . \infty$
iii. $(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\ldots \ldots \ldots \infty$
iv. $(1-x)^{-2}=1+2 x+3 x^{2}+4 x^{3}+\ldots \ldots \ldots \infty$
- 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 $\dfrac{1}{\mathrm{x}}$, which then will be small.
4. Exponential series
-
$\mathrm{e}^{\mathrm{x}}=1+\dfrac{\mathrm{x}}{1 !}+\dfrac{\mathrm{x}^{2}}{2 !}+\dfrac{\mathrm{x}^{3}}{3 !}+\ldots . \infty$
-
$\mathrm{e}=1+\dfrac{1}{1 !}+\dfrac{1}{2 !}+\dfrac{1}{3 !}+\ldots . . \infty(\mathrm{e} \simeq 2.72)$
-
$\mathrm{e}+\mathrm{e}^{-1}=2\left(1+\dfrac{1}{2 !}+\dfrac{1}{4 !}+\dfrac{1}{6 !}+\ldots . . \infty\right)$
-
$\mathrm{e}-\mathrm{e}^{-1}=2\left(\dfrac{1}{1 !}+\dfrac{1}{3 !}+\dfrac{1}{5 !}+\dfrac{1}{7 !}+\ldots . . \infty\right)$
5. Logarithmic series
For $-1<\mathrm{x} \leq 1$
$\log _{e}(1+x)=x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}-\dfrac{x^{4}}{4}+\ldots . \infty$
-
$\log _{e}(1-x)=-x-\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}-\dfrac{x^{4}}{4}+\ldots . \infty,-1 \leq x<1$
-
$\log \left(\dfrac{1+x}{1-x}\right)=2\left(x+\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}+\ldots . \infty\right),-1<x<1$
-
$\log _{\mathrm{e}} 2=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\ldots . . \infty \approx 0.693$
Solved Examples
1. Let $(1+\mathrm{x})^{\mathrm{n}}=\sum\limits _{\mathrm{r}=0}^{\mathrm{n}} \mathrm{a} _{\mathrm{r}} \mathrm{x}^{\mathrm{r}}$, then $\left(1+\dfrac{a _{1}}{a _{0}}\right)\left(1+\dfrac{a _{2}}{a _{1}}\right)\left(1+\dfrac{a _{3}}{a _{2}}\right) \ldots \ldots .\left(1+\dfrac{a _{n}}{a _{n-1}}\right)$ is equal to
(a) $\dfrac{(\mathrm{n}+1)^{\mathrm{n}+1}}{\mathrm{n} !}$
(b) $\dfrac{(\mathrm{n}+1)^{\mathrm{n}}}{\mathrm{n} !}$
(c) $\dfrac{(n)^{n-1}}{(n-1) !}$
(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.
$\begin{aligned} & \therefore\left(1+\dfrac{{ }^{n} C _{1}}{{ }^{n} C _{0}}\right)\left(1+\dfrac{{ }^{n} C _{2}}{{ }^{n} C _{1}}\right)\left(1+\dfrac{{ }^{n} C _{3}}{{ }^{n} C _{2}}\right) \cdots \cdots \cdot\left(1+\dfrac{{ }^{n} C _{n}}{{ }^{n} C _{n-1}}\right) \\ & =\left(1+\dfrac{n-0}{1}\right)\left(1+\dfrac{n-1}{2}\right)\left(1+\dfrac{n-2}{3}\right) \ldots \ldots .\left(1+\dfrac{n-n+1}{n}\right) \\ & =\dfrac{1+n}{1} \cdot \dfrac{1+n}{2} \cdot \dfrac{1+n}{3} \cdots \cdots \cdot \dfrac{1+n}{n}=\dfrac{(n+1)^{n}}{n !} \end{aligned}$
Answer b
2. If $\mathrm{a} _{\mathrm{n}}=\sum\limits _{\mathrm{r}=0}^{\mathrm{n}} \dfrac{1}{{ }^{n} \mathrm{C} _{\mathrm{r}}}$, then $\sum\limits _{\mathrm{r}=0}^{\mathrm{n}} \dfrac{\mathrm{r}}{{ }^{n} \mathrm{C} _{\mathrm{r}}}$ equals
(a) $(n-1) \cdot a _{n}$
(b) $n _{n}$
(c) $\dfrac{1}{2} \mathrm{na} _{\mathrm{n}}$
(d) None of these
Show Answer
Solution:
Let $S=\sum\limits _{\mathrm{r}=0}^{\mathrm{n}} \dfrac{\mathrm{r}}{{ }^{n} \mathrm{C} _{\mathrm{r}}}$
$\Rightarrow S=\dfrac{0}{{ }^{n} C _{0}}+\dfrac{1}{{ }^{n} C _{1}}+\dfrac{2}{{ }^{n} C _{2}}+\dfrac{3}{{ }^{n} C _{3}}+\ldots \ldots \ldots+\dfrac{n}{{ }^{n} C _{n}}$
Also, $S=\dfrac{n}{{ }^{n} C _{0}}+\dfrac{n-1}{{ }^{n} C _{1}}+\dfrac{n-2}{{ }^{n} C _{2}}+\dfrac{n-3}{{ }^{n} C _{3}}+\ldots \ldots \ldots+\dfrac{0}{{ }^{n} C _{n}}$
Adding (1) and 2
$\begin{aligned} & 2 S=\dfrac{n}{{ }^{n} C _{0}}+\dfrac{n}{{ }^{n} C _{1}}+\dfrac{n}{{ }^{n} C _{2}}+\ldots \ldots .+\dfrac{n}{{ }^{n} C _{n}} \\ & 2 S=n\left(\dfrac{1}{{ }^{n} C _{0}}+\dfrac{1}{{ }^{n} C _{1}}+\dfrac{1}{{ }^{n} C _{2}}+\ldots \ldots .+\dfrac{1}{{ }^{n} C _{n}}\right) \\ & \Rightarrow 2 S=n a _{n} \Rightarrow S=\dfrac{1}{2} n a _{n} \end{aligned}$
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 $\mathrm{x}=-\mathrm{i}$
$\Rightarrow(1+i)^{10}=\left(a _{0}-a _{2}+a _{4}-a _{6}+a _{8}-a _{10}\right)+i\left(a _{1}-a _{3}+a _{5}+a _{5}-a _{7}+a _{9}\right)$
Also, $(1-i)^{10}=\left(a _{0}-a _{2}+a _{4}-a _{6}+a _{8}-a _{10}\right)-i\left(a _{1}-a _{3}+a _{5}-a _{7}+a _{9}\right)$
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 $=(1-x)^{5}\left(1+x+x^{2}+x^{3}\right)^{4}=(1-x)^{5}\left((1+x)\left(1+x^{2}\right)\right)^{4}$
$=(1-\mathrm{x})\left\{(1-\mathrm{x})(1+\mathrm{x})\left(1+\mathrm{x}^{2}\right)\right\}^{4}$
$=(1-\mathrm{x})\left\{\left(1-\mathrm{x}^{4}\right)^{4}\right.$
$=(1-x)\left(1-4 x^{4}+6 x^{8}-4 x^{12}+x^{16}\right)$
$\therefore \quad$ 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) $2^{20}+\dfrac{20 !}{(10 !)^{2}}$
(b) $2^{19}-\dfrac{1}{2} \dfrac{20 !}{(10 !)^{2}}$
(c) $2^{19}+\dfrac{20 !}{(10 !)^{2}}$
(d) none of these
Show Answer
Solution:
In the expansion of $(1+\mathrm{x})^{20}$, put $\mathrm{x}=1$
$\begin{aligned} & 2^{20}={ }^{20} \mathrm{C} _{0}+{ }^{20} \mathrm{C} _{1}+{ }^{20} \mathrm{C} _{2}+\ldots \ldots \ldots \ldots \ldots+{ }^{20} \mathrm{C} _{9}+{ }^{20} \mathrm{C} _{10}+\ldots \ldots .{ }^{20} \mathrm{C} _{20} \\ & =2\left({ }^{20} \mathrm{C} _{0}+{ }^{+20} \mathrm{C} _{1}+\ldots \ldots \ldots \ldots \ldots+{ }^{20} \mathrm{C} _{10}\right)-{ }^{20} \mathrm{C} _{10} \quad\left(\because{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}-\mathrm{r}}\right) \\ & \Rightarrow \dfrac{2^{20}+{ }^{20} \mathrm{C} _{10}}{2}={ }^{20} \mathrm{C} _{0}+{ }^{20} \mathrm{C} _{1}+\ldots \ldots \ldots \ldots . .+{ }^{20} \mathrm{C} _{10} \\ & =2^{19}+\dfrac{1}{2} \dfrac{20 !}{(10 !)^{2}}={ }^{20} \mathrm{C} _{0}+{ }^{20} \mathrm{C} _{1}+\ldots \ldots \ldots \ldots .+{ }^{20} \mathrm{C} _{10} \end{aligned}$
Answer d
Exercise
1. The coefficient of $\mathrm{x}^{4}$ in $\left(\dfrac{\mathrm{x}}{2}-\dfrac{3}{\mathrm{x}^{2}}\right)^{10}$ is
(a) $\dfrac{405}{256}$
(b) $\dfrac{405}{259}$
(c) $\dfrac{450}{263}$
(d) None of these
Show Answer
Answer: a2. 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
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Answer: b3. The sum $\sum\limits _{i=1}^{m}\left(\begin{array}{c}10 \ i\end{array}\right)\left(\begin{array}{c}20 \ m-i\end{array}\right)$, 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: c4. 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) $\quad{ }^{12} \mathrm{C} _{6}+3$
(b) $\quad{ }^{12} \mathrm{C} _{6}+1$
(c) $\quad{ }^{12} \mathrm{C} _{6}$
(d) $\quad{ }^{12} \mathrm{C} _{6}+2$
Show Answer
Answer: d5. 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) $(\sqrt{3}, 2)$
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Answer: d6. If $(1+a x)^{n}=1+8 x+24 x^{2}$ then $\mathrm{a}=\ldots$ and $\mathrm{n}=$ ____________.
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Answer: a=2, n=47. The greatest term in the expansion of $\sqrt{3}\left(1+\dfrac{1}{\sqrt{3}}\right)^{20}$ is
(a) $\quad\left(\begin{array}{c}20 \ 7\end{array}\right) \dfrac{1}{27}$
(b) $\quad\left(\begin{array}{c}20 \ 6\end{array}\right) \dfrac{1}{81}$
(c) $\quad \dfrac{1}{9}\left(\begin{array}{c}20 \ 9\end{array}\right)$
(d) none of these
Show Answer
Answer: a8. 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: b9. 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
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Answer: c10. If $\dfrac{1}{(1-a x)(1-b x)}=a _{0}+a _{1}+a _{2} x^{2} \ldots .$. , then $a _{n}=$
(a) $\dfrac{a^{n+1}-b^{n+1}}{b-a}$
(b) $\dfrac{b^{n+1}-a^{n+1}}{b-a}$
(c) $\dfrac{b^{n}-a^{n}}{b-a}$
(d) $\dfrac{a^{n}-b^{n}}{b-a}$
Show Answer
Answer: b11. 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 $\left(a _{1}+a _{2}+a _{3}+\ldots . .+a _{m}\right)^{n}=\sum \dfrac{n !}{n _{1} ! n _{2} ! n _{3} ! \ldots \ldots . n _{m} !} \cdot a _{1}{ }^{n _{1}} \cdot a _{2}{ }^{n _{2}} \cdot a _{3}{ }^{n _{3}} \ldots \ldots . a _{m}{ }^{n _{m}}$ 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) ${ }^{n+2} \mathrm{C} _{4}$
(c) ${ }^{n+3} C _{4}$
(d) ${ }^{n+4} \mathrm{C} _{4}$
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
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
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) $5.2^{23}$
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: (i) c, (ii) c (iii) b (iv) c (v) b12.* Match the following:
| Column I | Column II | ||
|---|---|---|---|
| (a) | If $n$ be the degree of the polynomial $\sqrt{\left(3 x^2+1\right)}\left(\left(x+\sqrt{\left(3 x^2+1\right)}\right)^7-\left(x-\sqrt{\left(3 x^2+1\right)}\right)^2\right),$ 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 $n$ is divisible by | (q) | 4 |
| (c) | The sum of the binomial coefficients in the expansion of $\left(x^{-3 / 4}+n x^{5 / 4}\right)^m$, where $m$ 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 |
