Solution

Given expansion is $(\frac{3 x^{2}}{2}-\frac{1}{3 x})^{15}$

Let $T_{r+1}$ term is the general term.

Then,

$ \begin{aligned} T_{r+1} & ={ }^{15} C_{r}({\frac{3 x^{2}}{2}})^{15-r}(-\frac{1}{3 x})^{r} \\ & ={ }^{15} C_{r} 3^{15-r} x^{30-2 r} 2^{r-15}(-1)^{r} \cdot 3^{-r} \cdot x^{-r} \\ & ={ }^{15} C_{r}(-1)^{r} 3^{15-2 r} 2^{r-15} x^{30-3 r} \end{aligned} $

For independent of $x$,

$ \begin{aligned} 30-3 r & =0 \\ 3 r & =30 \Rightarrow r=10 \\ \because \quad T_{r+1} & =T_{10+1}=11 \text { th term is independent of } x . \\ \therefore \quad T_{10+1} & ={ }^{15} C_{10}(-1)^{10} 3^{15-20} 2^{10-15} \\ & ={ }^{15} C_{10} 3^{-5} 2^{-5} \\ & ={ }^{15} C_{10}(6)^{-5} \\ & ={ }^{15} C_{10} (\frac{1}{6})^5 \end{aligned} $

Solution

Given expansion is $(\sqrt{x}-{\frac{k}{x^{2}}})^{10}$.

Let $T_{r+1}$ is the general term.

Then,

$ \begin{aligned} T_{r+1} & ={ }^{10} C_{r}(\sqrt{x})^{10-r} (\frac{-k}{x^{2}})^{r} \\ & ={ }^{10} C_{r}(x)^{\frac{1}{2}(10-r)}(-k)^{r} \cdot x^{-2 r} \\ & ={ }^{10} C_{r} x^{5-\frac{r}{2}}(-k)^{r} \cdot x^{-2 r} \\ & ={ }^{10} C_{r} x^{5-\frac{r}{2}-2 r}(-k)^{r} \\ & ={ }^{10} C_{r} x^{\frac{10-5 r}{2}}(-k)^{r} \end{aligned} $

For free from $x, \quad \frac{10-5 r}{2}=0$

$\Rightarrow \quad 10-5 r=0 \Rightarrow r=2$

Since, $T_{2+1}=T_3$ is free from $x$.

$ \begin{aligned} & \therefore & T_{2+1} & ={ }^{10} C_2(-k)^{2}=405 \\ \Rightarrow & & \frac{10 \times 9 \times 8 !}{2 ! \times 8 !}(-k)^{2} & =405 \\ \Rightarrow 45 k^{2} & =405 \Rightarrow & k^{2}=\frac{405}{45}=9 \\ & \therefore & k & = \pm 3 \end{aligned} $

Solution

Given, $\quad$ expansion $=(1-3 x+7 x^{2})(1-x)^{16}$.

$ \begin{aligned} & =(1-3 x+7 x^{2})({ }^{16} C_0 1^{16}-{ }^{16} C_1 1^{15} x^{1}+{ }^{16} C_2 1^{14} x^{2}+\ldots+{ }^{16} C_{16} x^{16}) \\ & =(1-3 x+7 x^{2})(1-16 x+120 x^{2}+\ldots) \end{aligned} $

$\therefore \quad$ Coefficient of $x=-3-16=-19$

Solution

Given expansion is $(3 x-{\frac{2}{x^{2}}})^{15}$.

Let $T_{r+1}$ is the general term.

$ \begin{aligned} \therefore T_{r+1} & ={ }^{15} C_{r}(3 x)^{15-r} \frac{-2}{x^{2}}={ }^{r}{ }^{15} C_{r}(3 x)^{15-r}(-2)^{r} x^{-2 r} \\ & ={ }^{15} C_{r} 3^{15-r} x^{15-3 r}(-2)^{r} \end{aligned} $

For independent of $x, \quad 15-3 r=0 \Rightarrow r=5$

Since, $T_{5+1}=T_6$ is independent of $x$.

$ \begin{aligned} \therefore \quad T_{5+1} & ={ }^{15} C_5 3^{15-5}(-2)^{5} \\ & =-\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 !}{5 \times 4 \times 3 \times 2 \times 1 \times 10 !} \cdot 3^{10} \cdot 2^{5} \\ & =-3003 \cdot 3^{10} \cdot 2^{5} \end{aligned} $

Solution

(i) Given expansion is $(\frac{x}{a}-\frac{a}{x})^{10}$

Here, the power of Binomial i.e., $n=10$ is even.

Since, it has one middle term $(\frac{10}{2}+1)$ th term i.e., 6th term.

$ \begin{aligned} \therefore \quad T_6 & =T_{5+1}={ }^{10} C_5 (\frac{x}a)^{10-5} (\frac{-a}{x})^5 \\ & =-{ }^{5} C_5 (\frac{x}a)^{5} (\frac{a}x)^{5} \\ & =-\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2 \times 1} \quad (\frac{x}{a})^5 \quad (\frac{x}{a})^{-5} \\ & =-9 \times 4 \times 7=-252 \end{aligned} $

(ii) Given expansion is $(3 x-{\frac{x^{3}}{6}})^{9}$.

Here, $n=9 \quad$ [odd]

Since, the Binomial expansion has two middle terms i.e., $(\frac{9+1}{2})$ th and $(\frac{9+1}{2}+1)$ th i.e., 5 th term and 6 th term.

$ \begin{aligned} \therefore \quad T_5 & =T_{(4+1)}={ }^{9} C_4(3 x)^{9-4}(-\frac{x^{3}}{6})^4 \\ & =\frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !} 3^{5} x^{5} x^{12} 6^{-4} \\ & =\frac{7 \times 6 \times 3 \times 3^{1}}{2^{4}} x^{17}=\frac{189}{8} x^{17} \end{aligned} $

$ \begin{aligned} \therefore \quad T_6 & =T_{5+1}={ }^{9} C_5(3 x)^{9-5}(-\frac{x^{3}}{6})^5 \\ & =-\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1} \cdot 3^{4} \cdot x^{4} \cdot x^{15} \cdot 6^{-5} \\ & =\frac{-21 \times 6}{3 \times 2^{5}} x^{19}=\frac{-21}{16} x^{19} \end{aligned} $

Solution

Given expansion is $(x-x^{2})^{10}$.

Let the term $T_{r+1}$ is the general term.

$ \begin{aligned} \therefore \quad T_{r+1} & ={ }^{10} C_{r} x^{10-r}(-x^{2})^{r} \\ & =(-1)^{r} \cdot{ }^{10} C_{r} \cdot x^{10-r} \cdot x^{2 r} \\ & =(-1)^{1}{ }^{10} C_{r} x^{10+r} \end{aligned} $

For the coefficient of $x^{15}$,

$ \begin{aligned} 10+r & =15 \Rightarrow r=5 \\ T_{5+1} & =(-1)^{5}{ }^{10} C_5 x^{15} \\ \therefore \quad \text { Coefficient of } x^{15} & =-1 \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 \times 4 \times 3 \times 2 \times 1 \times 5 !} \\ & =-3 \times 2 \times 7 \times 6=-252 \end{aligned} $

Solution

Given expansion is $(x^{4}-{\frac{1}{x^{3}}})^{15}$.

Let the term $T_{r+1}$ contains the coefficient of $\frac{1}{x^{17}}$ i.e., $x^{-17}$.

$ \therefore \quad \begin{aligned} T_{r+1} & ={ }^{15} C_{r}(x^{4})^{15-r}(-\frac{1}{x^{3}}{ })^{r} \\ & ={ }^{15} C_{r} x^{60-4 r}(-1)^{r} x^{-3 r} \\ & ={ }^{15} C_{r} x^{60-7 r}(-1)^{r} \end{aligned} $

For the coefficient $x^{-17}$,

$ \begin{aligned} & 60-7 r=-17 \\ & \Rightarrow \quad 7 r=77 \Rightarrow r=11 \\ & \Rightarrow \quad T_{11+1}={ }^{15} C_{11} x^{60-77}(-1)^{11} \\ & \therefore \quad \text { Coefficient of } x^{-17}=\frac{-15 \times 14 \times 13 \times 12 \times 11 !}{11 ! \times 4 \times 3 \times 2 \times 1} \\ & =-15 \times 7 \times 13=-1365 \end{aligned} $

Solution

Given expansion is $(y^{1 / 2}+x^{1 / 3})^{n}$.

The sixth term of this expansion is

$ T_6=T_{5+1}={ }^{n} C_5(y^{1 / 2})^{n-5}(x^{1 / 3})^{5} \quad \ldots (i) $

Now, given that the Binomial coefficient of the third term from the end is 45 .

We know that, Binomial coefficient of third term from the end =Binomial coefficient of third term from the begining $={ }^{n} C_2$

$ \begin{aligned} & \because \quad{ }^{n} C_2=45 \\ & \Rightarrow \quad \frac{n(n-1)(n-2) !}{2 !(n-2) !}=45 \\ & \Rightarrow \quad n(n-1)=90 \\ & \Rightarrow \quad n^{2}-n-90=0 \\ & \Rightarrow \quad n^{2}-10 n+9 n-90=0 \\ & \Rightarrow \quad n(n-10)+9(n-10)=0 \\ & \Rightarrow \quad(n-10)(n+9)=0 \\ & \Rightarrow \quad(n+9)=0 \text { or }(n-10)=0 \\ & \therefore \quad n=10 \quad[\because n \neq-9] \end{aligned} $

From Eq. (i),

$ T_6={ }^{10} C_5 y^{5 / 2} x^{5 / 3}=252 y^{5 / 2} \cdot x^{5 / 3} $

Solution

Given expansion is $(1+x)^{18}$.

Now, $(2 r+4)$ th term i.e., $T_{2 r+3+1}$.

$ \begin{aligned} \therefore \quad T_{2 r+3+1} & ={ }^{18} C_{2 r+3}(1)^{18-2 r-3}(x)^{2 r+3} \\ & ={ }^{18} C_{2 r+3} x^{2 r+3} \end{aligned} $

Now, $\quad(r-2)$ th term i.e., $T_{r-3+1}$.

$ \therefore T_{r-3+1} =^{18} C_{r-3} x^{r-3} $

$ \text { As, } ^{18} C_{2 r+3} =^{18} C_{r-3} $ $[\because ^nC_x = ^nC_y \Rightarrow x+y=n]$

$ \Rightarrow 2 r+3+r-3 =18 $

$ \Rightarrow 3 r =18 $

$ \therefore r =6 $

Solution

Given expansion is $(1+x)^{2 n}$.

Now, coefficient of 2nd term $={ }^{2 n} C_1$

Coefficient of 3rd term $={ }^{2 n} C_2$

Coefficient of 4 th term $={ }^{2 n} C_3$

Given that, ${ }^{2 n} C_1,{ }^{2 n} C_2$ and ${ }^{2 n} C_3$ are in AP.

Then, $\quad 2 \cdot{ }^{2 n} C_2={ }^{2 n} C_1+{ }^{2 n} C_3$

$\Rightarrow \quad 2 [\frac{2 n(2 n-1)(2 n-2) !}{2 \times 1 \times(2 n-2) !}]=\frac{2 n(2 n-1) !}{(2 n-1) !}+\frac{2 n(2 n-1)(2 n-2)(2 n-3) !}{3 !(2 n-3) !}$

$\Rightarrow \quad n(2 n-1)=n+\frac{n(2 n-1)(2 n-2)}{6}$

$\Rightarrow \quad n(12 n-6)=n(6+4 n^{2}-4 n-2 n+2)$

$\Rightarrow \quad 12 n-6=(4 n^{2}-6 n+8)$

$\Rightarrow \quad 6(2 n-1)=2(2 n^{2}-3 n+4)$

$\Rightarrow \quad 3(2 n-1)=2 n^{2}-3 n+4$

$\Rightarrow \quad 2 n^{2}-3 n+4-6 n+3=0$

$\Rightarrow \quad 2 n^{2}-9 n+7=0$

Solution

Given, expansion $=(1+x+x^{2}+x^{3})^{11}=[(1+x)+x^{2}(1+x)]^{11}$

$ =[(1+x)(1+x^{2})]^{11}=(1+x)^{11} \cdot(1+x^{2})^{11} $

Now, above expansion becomes

$ \begin{aligned} & =({ }^{11} C_0+{ }^{11} C_1 x+{ }^{11} C_2 x^{2}+{ }^{11} C_3 x^{3}+{ }^{11} C_4 x^{4}+\ldots)({ }^{11} C_0+{ }^{11} C_1 x^{2}+{ }^{11} C_2 x^{4}+\ldots) \\ & =(1+11 x+55 x^{2}+165 x^{3}+330 x^{4}+\ldots)(1+11 x^{2}+55 x^{4}+\ldots) \end{aligned} $

$\therefore$ Coefficient of $x^{4}=55+605+330=990$

Solution

Given expansion is $(\frac{p}{2}+2)^{8}$.

Here, $n=8 \quad$ [even]

Since, this Binomial expansion has only one middle term i.e., $(\frac{8}{2}+1)$ th $=5$ th term

$ T_5 =T_{4+1}={ }^{8} C_4 (\frac{p}2)^{8-4} \cdot 2^{4} $

$ \Rightarrow 1120 ={ }^{8} C_4 p^{4} \cdot 2^{-4} 2^{4} $

$ \Rightarrow 1120 =\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 4 \times 3 \times 2 \times 1} p^{4} $

$ \begin{matrix} \Rightarrow & 1120=7 \times 2 \times 5 \times p^{4} \\ \Rightarrow & p^{4}=\frac{1120}{70}=16 \Rightarrow p^{4}=2^{4} \\ \Rightarrow & p^{2}=4 \Rightarrow p= \pm 2 \end{matrix} $

Solution

Given, expansion is $(x-\frac{1}x)^{2 n}$. This Binomial expansion has even power. So, this has one middle term.

i.e., $\begin{aligned} (\frac{2 n}{2}+1) & \text { th term }=(n+1) \text { th term } \\ T_{n+1} & ={ }^{2 n} C_{n}(x)^{2 n-n}(-\frac{1}{x}{ })^{n}={ }^{2 n} C_{n} x^{n}(-1)^{n} x^{-n} \\ & ={ }^{2 n} C_{n}(-1)^{n}=(-1)^{n} \frac{(2 n) !}{n ! n !}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots(2 n-1)(2 n)}{n ! n !}(-1)^{n} \\ & =\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1) \cdot 2 \cdot 4 \cdot 6 \ldots(2 n)}{12 \cdot 3 \cdot \ldots n(n !)}(-1)^{n} \\ & =\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1) \cdot 2^{n}(1 \cdot 2 \cdot 3 \ldots n)(-1)^{n}}{(1 \cdot 2 \cdot 3 \ldots n)(n !)} \\ & =\frac{[1 \cdot 3 \cdot 5 \ldots(2 n-1)]}{n !}(-2)^{n} \end{aligned} $ Hence proved.

Solution

Here, the Binomial expansion is $(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}{ })^{n}$.

Now, 7th term from beginning $T_7=T_{6+1}={ }^{n} C_6(\sqrt[3]{2})^{n-6} (\frac{1}{\sqrt[3]{3}})^{6} \quad \ldots (i)$

and 7 th term from end i.e., $T_7$ from the beginning of $(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{2}{ })^{n}$

i.e., $\quad T_7={ }^{n} C_6 (\frac{1}{\sqrt[3]{3}})^{n-6}(\sqrt[3]{2})^{6}$

Given that, $\frac{{ }^{n} C_6(\sqrt[3]{2})^{n-6} (\frac{1}{\sqrt[3]{3}})^6}{{ }^{n} C_6 (\frac{1}{\sqrt[3]{3}})^{n-6}(\sqrt[3]{2})^{6}}=\frac{1}{6} \Rightarrow \frac{2^{\frac{n-6}{3}} \cdot 3^{-6 / 3}}{3^{-\frac{n-6}{3}} \cdot 2^{6 / 3}}=\frac{1}{6}$

$\Rightarrow \quad( 2^{\frac{n-6}{3}} \cdot 2^{\frac{-6}{3}}) (3^{\frac{-6}{3}} \cdot 3^{\frac{(n-6)}{3}})=6^{-1}$

$ \begin{aligned} & \Rightarrow \quad 2^{\frac{n-6}{3}-\frac{6}{3}} \cdot 3^{\frac{n-6}{3}-\frac{6}{3}}=6^{-1} \Rightarrow(2 \cdot 3)^{\frac{n}{3}-4}=6^{-1} \\ & \Rightarrow \quad \frac{n}{3}-4=-1 \Rightarrow \frac{n}{3}=3 \\ & \therefore \quad n=9 \end{aligned} $

Solution

(i) Given expansion is $(x+a)^{n}$.

$\therefore(x+a)^{n}={ }^{n} C_0 x^{n} a^{0}+{ }^{n} C_1 x^{n-1} a^{1}+{ }^{n} C_2 x^{n-2} a^{2}+{ }^{n} C_3 x^{n-3} a^{3}+\ldots+{ }^{n} C_{n} a^{n}$

Now, sum of odd terms

i.e.,

$ O={ }^{n} C_0 x^{n}+{ }^{n} C_2 x^{n-2} a^{2}+\ldots $

and sum of even terms

i.e.,

$ E={ }^{n} C_1 x^{n-1} a+{ }^{n} C_3 x^{n-3} a^{3}+\ldots $

$ \begin{aligned} & \because \quad(x+a)^{n}=O+E \\ & \text { Similarly, } \quad(x-a)^{n}=O-E \\ & \therefore \quad(O+E)(O-E)=(x+a)^{n}(x-a)^{n} \quad \text { [on multiplying Eqs. (i) and (ii)] } \\ & \Rightarrow \quad O^{2}-E^{2}=(x^{2}-a^{2})^{n} \end{aligned} $

(ii) $4 O E=(O+E)^{2}-(O-E)^{2}=[(x+a)^{n}]^{2}-[(x-a)^{n}]^{2}$

[from Eqs. (i) and (ii)]

$ =(x+a)^{2 n}-(x-a)^{2 n} $

Hence proved.

Solution

Given expansion is $x^{2}+\frac{1}x^{2 n}$.

Let $x^{p}$ occur in the expansion of $x^{2}+\frac{1}x^{2 n}$.

$ \begin{aligned} T_{r+1} & ={ }^{2 n} C_{r}(x^{2})^{2 n-r} \frac{1}{x} \\ & ={ }^{2 n} C_{r} x^{4 n-2 r} x^{-r}={ }^{2 n} C_{r} x^{4 n-3 r} \\ \text { Let } \quad 4 n-3 r & =p \\ \Rightarrow \quad 3 r & =4 n-p \Rightarrow r=\frac{4 n-p}{3} \\ \therefore \quad \text { Coefficient of } x^{p} & ={ }^{2 n} C_{r}=\frac{(2 n) !}{r !(2 n-r) !}=\frac{(2 n) !}{\frac{4 n-p}{3} ! 2 n-\frac{4 n-p}{3} !} \\ & =\frac{(2 n) !}{\frac{4 n-p}{3} ! \frac{6 n-4 n+p}{3} !}=\frac{(2 n) !}{\frac{4 n-p}{3} ! \frac{2 n+p}{3} !} \end{aligned} $

Solution

Given expansion is $(1+x+2 x^{3}) \frac{3}{2} x^{2}-\frac{1}{3 x}^{9}$.

Now, consider $\frac{3}{2} x^{2}-\frac{1}{3 x}^{9}$

$ \begin{aligned} T_{r+1} & ={ }^{9} C_{r} \frac{3}{2} x^{2}-\frac{1}{3 x} \\ & ={ }^{9} C_{r} \frac{3}2^{9-r} x^{18-2 r}-\frac{1}3^{r} x^{-r}={ }^{9} C_{r} \frac{3}2^{9-r}-\frac{1}3^{r} x^{18-3 r} \end{aligned} $

Hence, the general term in the expansion of $(1+x+2 x^{3}) \frac{3}{2} x^{2}-\frac{1}{3 x}^{9}$

$ ={ }^{9} C_{r} \frac{3}2^{9-r}-\frac{1}3^{r} x^{18-3 r}+{ }^{9} C_{r} \frac{3}2^{9-r}-\frac{1}3^{r} x^{19-3 r}+2 \cdot{ }^{9} C_{r} \frac{3}2^{9-r}-\frac{1}3^{r} x^{21-3 r} $

For term independent of $x$, putting $18-3 r=0,19-3 r=0$ and $21-3 r=0$, we get

$ r=6, r=19 / 3, r=7 $

Since, the possible value of $r$ are 6 and 7 .

Hence, second term is not independent of $x$.

$\therefore \quad$ The term independent of $x$ is ${ }^{9} C_6 \frac{3}2^{9-6}-\frac{1}3^{6}+2 \cdot{ }^{9} C_7 \frac{3}2^{9-7}-\frac{1}3^{7}$

$ \begin{aligned} & =\frac{9 \times 8 \times 7 \times 6 !}{6 ! \times 3 \times 2} \cdot \frac{3^{3}}{2^{3}} \cdot \frac{1}{3^{6}}-2 \cdot \frac{9 \times 8 \times 7 !}{7 ! \times 2 \times 1} \cdot \frac{3^{2}}{2^{2}} \cdot \frac{1}{3^{7}} \\ & =\frac{84}{8} \cdot \frac{1}{3^{3}}-\frac{36}{4} \cdot \frac{2}{3^{5}}=\frac{7}{18}-\frac{2}{27}=\frac{21-4}{54}=\frac{17}{54} \end{aligned} $

Solution

(c) Here, $(x+a)^{100}+(x-a)^{100}$

Total number of terms is 102 in the expansion of $(x+a)^{100}+(x-a)^{100}$

50 terms of $(x+a)^{100}$ cancel out 50 terms of $(x-a)^{100} .51$ terms of $(x+a)^{100}$ get added to the 51 terms of $(x-a)^{100}$.

Alternate Method

$ \begin{aligned} (x+a)^{100}+(x-a)^{100} & ={ }^{100} C_0 x^{100}+{ }^{100} C_1 x^{99} a+\ldots+{ }^{100} C_{100} a^{100} \\ & +{ }^{100} C_0 x^{100}-{ }^{100} C_1 x^{99} a+\ldots+{ }^{100} C_{100} a^{100} \\ & =2 \underbrace{.{ }^{100} C_0 x^{100}+{ }^{100} C_2 x^{98} a^{2}+\ldots+{ }^{100} C_{100} a^{100}]}_{51 \text { terms }} \end{aligned} $

Solution

(a) Given that, $r>1, n>2$ and the coefficients of $(3 r)$ th and $(r+2)$ th term are equal in the expansion of $(1+x)^{2 n}$.

$\begin{aligned} & \text { Then, } \quad T_{3 r}=T_{3 r-1+1}={ }^{2 n} C_{3 r-1} x^{3 r-1} \\ & \text { and } \quad T_{r+2}=T_{r+1+1}={ }^{2 n} C_{r+1} x^{r+1} \\ & \text { Given, } \quad{ }^{2 n} C_{3 r-1}={ }^{2 n} C_{r+1} \quad\left[\because{ }^n C_x={ }^n C_y \Rightarrow x+y=n\right] \\ & \Rightarrow \quad 3 r-1+r+1=2 n \\ & \Rightarrow \quad 4 r=2 n \Rightarrow n=\frac{4 r}{2} \\ & \therefore \quad n=2 r \\ & \end{aligned}$

Solution

(c) Let two successive terms in the expansion of $(1+x)^{24}$ are $(r+1)$ th and $(r+2)$ th terms.

$ \therefore $

and

$ T_{r+1} ={ }^{24} C_{r} x^{r} \\ $

$T_{r+2} ={ }^{24} C_{r+1} x^{r+1} \\ $

$\frac{{ }^{24} C_{r}}{{ }^{24} C_{r+1}} =\frac{1}{4} $

Given that,

$ \Rightarrow \frac{\frac{(24) !}{r !(24-r) !}}{(24) !} =\frac{1}{4} \\ $

$ \Rightarrow \frac{(r+1) !(24-r-1) !}{(r+1) r(23-r) !} =\frac{1}{4} \\ $

$\Rightarrow \frac{(r+124-r)(23-r) !}{r 4-r} =\frac{1}{4} \Rightarrow 4 r+4=24-r \\ $

$\Rightarrow 5 r=20 \Rightarrow r=4 \\ $

$ \therefore T_{4+1} =T_5 \text { and } T_{4+2}=T_6 $

Hence, 5th and 6th terms.

Solution

(d) $\because$ Coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n}={ }^{2 n} C_{n}$ and coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n-1}={ }^{2 n-1} C_{n}$

$ \frac{{ }^{2 n} C_{n}}{{ }^{2 n-1} C_{n}} =\frac{\frac{(2 n) !}{n ! n !}}{\frac{(2 n-1) !}{n !(n-1) !}} \\ $

$ =\frac{(2 n) ! n !(n-1) !}{n ! n !(2 n-1) !} \\ $

$ =\frac{2 n(2 n-1) ! n !(n-1) !}{n ! n(n-1) !(2 n-1) !} \\ $

$ =\frac{2 n}{n}=\frac{2}{1}=2: 1 $

Solution

(b) The expansion of $(1+x)^{n}$ is ${ }^{n} C_0+{ }^{n} C_1 x+{ }^{n} C_2 x^{2}+{ }^{n} C_3 x^{3}+\ldots+{ }^{n} C_{n} x^{n}$

$\therefore \quad$ Coefficient of 2 nd term $={ }^{n} C_1$,

Coefficient of 3rd term $={ }^{n} C_2$

and coefficient of 4 th term $={ }^{n} C_3$

Given that, ${ }^{n} C_1,{ }^{n} C_2$ and ${ }^{n} C_3$ are in AP

$ \begin{matrix} \therefore & 2{ }^{n} C_2={ }^{n} C_1+{ }^{n} C_3 \\ \Rightarrow & 2 \frac{(n) !}{(n-2) ! 2 !}=\frac{(n) !}{(n-1) !}+\frac{(n) !}{3 !(n-3) !} \\ \Rightarrow & \frac{2 \cdot n(n-1)(n-2) !}{(n-2) ! 2 !}=\frac{n(n-1) !}{(n-1) !}+\frac{n(n-1)(n-2)(n-3) !}{3 \cdot 2 \cdot 1(n-3) !} \\ \Rightarrow & n(n-1)=n+\frac{n(n-1)(n-2)}{6} \\ \Rightarrow & 6 n-6=6+n^{2}-3 n+2 \\ \Rightarrow & n^{2}-9 n+14=0 \\ \Rightarrow & n(n-7)-2(n-7)=0 \\ \Rightarrow & (n-7)(n-2)=0 \\ \Rightarrow & n=2 \text { or } n=7 \end{matrix} $

Since, $n=2$ is not possible.

$ \therefore \quad n=7 $

Solution

(b) Since, the coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n}$ is ${ }^{2 n} C_{n}$.

$\therefore \quad A={ }^{2 n} C_{n}$

Now, the coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n-1}$ is ${ }^{2 n-1} C_{n}$.

$\therefore \quad B={ }^{2 n-1} C_{n}$

Now,

$ \frac{A}{B}=\frac{{ }^{2 n} C_{n}}{{ }^{2 n-1} C_{n}}=\frac{2}{1}=2 $

Same as solution No. 21.

Solution

(c) Given expansion is $\frac{1}{x}+x \sin x^{10}$.

Since, $n=10$ is even, so this expansion has only one middle term i.e., 6th term.

$ \begin{aligned} & \therefore \quad T_6=T_{5+1}={ }^{10} C_5 \frac{1}{x}{ }^{10-5}(x \sin x)^{5} \\ & \Rightarrow \quad \frac{63}{8}={ }^{10} C_5 x^{-5} x^{5} \sin ^{5} x \\ & \Rightarrow \quad \frac{63}{8}=\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 !}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 5 !} \sin ^{5} x \\ & \Rightarrow \quad \frac{63}{8}=2 \cdot 9 \cdot 2 \cdot 7 \cdot \sin ^{5} x \\ & \Rightarrow \quad \sin ^{5} x=\frac{1}{32} \\ & \Rightarrow \quad \sin ^{5} x=\frac{1}2^{5} \\ & \Rightarrow \quad \sin x=\frac{1}{2} \\ & \therefore \quad x=n \pi+(-1)^{n} \pi / 6 \end{aligned} $

Solution

Largest coefficient in the expansion of $(1+x)^{30}={ }^{30} C_{30 / 2}={ }^{30} C_{15}$

Solution

Given expansion is $(x+y+z)^{n}=[x+(y+z)]^{n}$.

$ [x+(y+z)]^{n}={ }^{n} C_0 x^{n}+{ }^{n} C_1 x^{n-1}(y+z) $

$ +{ }^{n} C_2 x^{n-2}(y+z)^{2}+\ldots+{ }^{n} C_{n}(y+z)^{n} $

$\therefore \quad$ Number of terms $=1+2+3+\ldots+n+(n+1)$

$ =\frac{(n+1)(n+2)}{2} $

Solution

Let constant be $T_{r+1}$.

$\therefore T_{r+1} ={ }^{16} C_{r}(x^{2})^{16-r}-\frac{1}{x^{2}} \\ $

$ ={ }^{16} C_{r} x^{32-2 r}(-1)^{r} x^{-2 r} \\ $ $ ={ }^{16} C_{r} x^{32-4 r}(-1)^{r} \\ $ $ \text { For constant term, } \quad 32-4 r =0 \Rightarrow r=8 \\ $ $\therefore T_{8+1} ={ }^{16} C_8 $

Solution

Given expansions is $\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}^{n}$

$ \therefore \quad \quad T_7=T_{6+1}={ }^{n} C_6(\sqrt[3]{2})^{n-6} \frac{1}{\sqrt[3]{3}}^{6} $

Since, $T_7$ from end is same as the $T_7$ from beginning of $\frac{1}{\sqrt[3]{3}}+\sqrt[3]{2}{ }^{n}$

Then,

$ T_7={ }^{n} C_6 \frac{1}{\sqrt[3]{3}}^{n-6}(\sqrt[3]{2})^{6} $

Given that,

$ { }^{n} C_6(2)^{\frac{n-6}{3}}(3)^{-6 / 3}={ }^{n} C_6(3)^{-\frac{(n-6)}{3}} 2^{6 / 3} $

$ \Rightarrow \quad \text { (2) } \frac{n-12}{3}={\frac{1}{3^{1 / 3}}}^{n-12} $

which is true, when $\frac{n-12}{3}=0$.

$\Rightarrow$

$ n-12=0 \Rightarrow n=12 $

Solution

Given expansion is $\frac{1}{a}-\frac{2 b}3^{10}$.

Let $T_{r+1}$ has the coefficient of $a^{-6} b^{4}$.

$\therefore \quad T_{r+1}={ }^{10} C_{r} \frac{1}a^{10-r}-\frac{2 b}3^{r}$

For coefficient of $a^{-6} b^{4}, \quad 10-r=6 \Rightarrow r=4$

Coefficient of $a^{-6} b^{4}={ }^{10} C_4(-2 / 3)^{4}$

$\therefore \quad=\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 !}{6 ! \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{2^{4}}{3^{4}}=\frac{1120}{27}$

Solution

Given expansion is $(a^{3}+b a)^{28}$.

$\because \quad n=28$

[even]

$\therefore \quad$ Middle term $=\frac{28}{2}+1$ th term $=15$ th term

$\therefore \quad T_{15}=T_{14+1}$

$={ }^{28} C_{14}(a^{3})^{28-14}(b a)^{14}$

$={ }^{28} C_{14} a^{42} b^{14} a^{14}$

$={ }^{28} C_{14} a^{56} b^{14}$

Solution

Given expansion is $(1+x)^{p+q}$.

$\therefore \quad$ Coefficient of $x^{p}={ }^{p+q} C_{P}$

and coefficient of $x^{q}={ }^{p+q} C_{q}$

$\therefore \quad \frac{{ }^{p+q} C_{p}}{{ }^{p+q} C_{q}}=\frac{{ }^{p+q} C_{p}}{{ }^{p+q} C_{p}}=1: 1$

Solution

Given expansion is $\sqrt{\frac{x}{3}}+{\frac{3}{2 x^{2}}}^{10}$.

Let the constant term be $T_{r+1}$.

Then,

$ \begin{aligned} T_{r+1} & ={ }^{10} C_{r} \sqrt{\frac{x}{3}}^{10-r}{\frac{3}{2 x^{2}}}^{r} \\ & ={ }^{10} C_{r} \cdot x^{\frac{10-r}{2}} \cdot 3^{\frac{-10+r}{2}} \cdot 3^{r} \cdot 2^{-r} \cdot x^{-2 r} \\ & ={ }^{10} C_{r} x^{\frac{10-5 r}{2}} 3^{\frac{-10+3 r}{2}} 2^{-r} \end{aligned} $

For constant term, $10-5 r=0 \Rightarrow r=2$

Hence, third term is independent of $x$.

Solution

Let

$ \begin{aligned} 25^{15} & =(26-1)^{15} \\ & ={ }^{15} C_0 26^{15}-{ }^{15} C_1 26^{14}+\ldots-{ }^{15} C_{15} \\ & ={ }^{15} C_0 26^{15}-{ }^{15} C_1 26^{14}+\ldots-1-13+13 \\ & ={ }^{15} C_0 26^{15}-{ }^{15} C_1 26^{14}+\ldots-13+12 \end{aligned} $

It is clear that, when $25^{15}$ is divided by 13 , then remainder will be 12 .

Solution

False

Given series

$ \begin{aligned} & =\sum_{r=0}^{10}{ }^{20} C_{r}={ }^{20} C_0+{ }^{20} C_1+{ }^{20} C_2+\ldots+{ }^{20} C_{10} \\ & ={ }^{20} C_0+{ }^{20} C_1+\ldots+{ }^{20} C_{10}+{ }^{20} C_{11}+\ldots{ }^{20} C_{20}-({ }^{20} C_{11}+\ldots+{ }^{20} C_{20}) \\ & =2^{20}-({ }^{20} C_{11}+\ldots+{ }^{20} C_{20}) \end{aligned} $

Hence, the given statement is false.

Solution

True

Given expression $=7^{9}+9^{7}=(1+8)^{7}-(1-8)^{9}$

$ \begin{aligned} & =({ }^{7} C_0+{ }^{7} C_1 8+{ }^{7} C_2 8^{2}+\ldots+{ }^{7} C_7 8^{7})-({ }^{9} C_0-{ }^{9} C_1 8+{ }^{9} C_2 8^{2} \ldots-{ }^{9} C_9 8^{9}) \\ & =(1+7 \times 8+21 \times 8^{2}+\ldots)-(1-9 \times 8+36 \times 8^{2}+\ldots-8^{9}) \\ & =(7 \times 8+9 \times 8)+(21 \times 8^{2}-36 \times 8^{2})+\ldots \\ & =2 \times 64+(21-36) 64+\ldots \end{aligned} $

which is divisible by 64

Hence, the statement is true.

Solution

False

Given expansion is $[(2 x+y^{3})^{4}]^{7}=(2 x+y^{3})^{28}$.

Since, this expansion has 29 terms.

So, the given statement is false.

Solution

False

Here, the Binomial expansion is $(1+x)^{2 n-1}$.

Since, this expansion has two middle term i.e., $\frac{2 n-1+1}{2}$ th term and $\frac{2 n-1+1}{2}+1$ th term i.e., $n$th term and $(n+1)$ th term.

$ \begin{aligned} \therefore \quad \text { Coefficient of } n \text {th term } & ={ }^{2 n-1} C_{n-1} \\ \text { Coefficient of }(n+1) \text { th term } & ={ }^{2 n-1} C_{n} \\ \text { Sum of coefficients } & ={ }^{2 n-1} C_{n-1}+{ }^{2 n-1} C_{n} \\ & ={ }^{2 n-1+1} C_{n}={ }^{2 n} C_{n} \quad[\because{ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}] \end{aligned} $

Solution

True

Given that, $\quad 3^{400}=9^{200}=(10-1)^{200}$

$ \begin{matrix} \Rightarrow & (10-1)^{200}={ }^{200} C_0 10^{200}-{ }^{200} C_1 10^{199}+\ldots-{ }^{200} C_{199} 10^{1}+{ }^{200} C_{200} 1^{200} \\ \Rightarrow & (10-1)^{200}=10^{200}-200 \times 10^{199}+\ldots-10 \times 200+1 \end{matrix} $

So, it is clear that the last two digits are 01 .

Solution

False

Given Binomial expansion is $x-{\frac{1}{x^{2}}}^{2 n}$.

Let $\quad T_{r+1}$ term is independent of $x$.

Then,

$ \begin{aligned} T_{r+1} & ={ }^{2 n} C_{r}(x)^{2 n-r}-{\frac{1}{x^{2}}}^{r} \\ & ={ }^{2 n} C_{r} x^{2 n-r}(-1)^{r} x^{-2 r}={ }^{2 n} C_{r} x^{2 n-3 r}(-1)^{r} \end{aligned} $

For independent of $x$,

$ \therefore \quad \begin{aligned} 2 n-3 r & =0 \\ r & =\frac{2 n}{3}, \end{aligned} $

which is not a integer.

So, the given expansion is not possible.

Solution

False

We know that, the number of terms in the expansion of $(a+b)^{n}$, where $n \in N$, is one more than the power $n$.