Exercise 5.1-Additional Questions - Chapter 5 - Binomial Theorem, Sequences and Series - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions Solved
Question 1.
Find the coefficient of the term involving $x^{32}$ and $x^{-17}$ in the expansion of $\left(x^4-\frac{1}{x^3}\right)^{15}$
Solution:
Let $T_{\mathrm{r}+1}$ be the term in which $\mathrm{x}^{32}$ and $\mathrm{x}^{-17}$ occurs,
$
\begin{aligned}
\therefore T_{r+1} & ={ }^{15} \mathrm{C}_r \cdot\left(x^4\right)^{15-r}\left(-\frac{1}{x^3}\right)^r \\
& ={ }^{15} \mathrm{C}_r,(-1)^r, x^{60-4 r}, x^{-3 r}={ }^{15} \mathrm{C}_r,(-1)^r, x^{60-7 r}
\end{aligned}
$
(i) Since $x^{32}$ occurs in this term
$\therefore$ Exponent of $\mathrm{x}=32$
$\Rightarrow 60-7 \mathrm{r}=32 \Rightarrow 7 \mathrm{r}=28$
$r=28 \div 7=4$
$\therefore$ Coefficient of the term containing $\mathrm{x}^{32}={ }^{15} \mathrm{C}_4(-1)^4=1365$
(ii) Since $x^{-17}$ occurs in this term .
$\therefore$ Exponent of $\mathrm{x}=-17$
$\Rightarrow 60-7 \mathrm{r}=-17 \Rightarrow 7 \mathrm{r}=77, \therefore \mathrm{r}=11$
$\therefore$ Coefficient of the term containing $x^{-17}={ }^{15} \mathrm{C}_{11}(-1)^{11}=-{ }^{15} \mathrm{C}_{15-11}=-{ }^{15} \mathrm{C}_4=-1365$
Question 2.
Find a positive value of $m$ for which the coefficient of $x^2$ in the expansion of $(1+x)^{\mathrm{m}}$ is 6 .
Solution:
The general term in the expansion of $(1+x)^m$ is $T_{r+1}={ }^m \mathrm{C}_r(1)^{m-r} x^r$
On putting $r=2$, we get $T_3={ }^m \mathrm{C}_2(1)^{m-2} x^2={ }^m \mathrm{C}_2 x^2$
$\therefore$ Coefficient of $x^2={ }^m \mathrm{C}_2$
Also, coefficient of $x^2$ in the expansion of $(1+x)^{\mathrm{m}}$ is 6
$
\begin{aligned}
& \therefore{ }^m C_2=6 \Rightarrow \frac{m(m-1)}{2.1}=6 \Rightarrow m(m-1)=12 \\
& \Rightarrow m(m-1)=4.3 \Rightarrow \mathrm{m}=4
\end{aligned}
$

Question 3.
In the binomial expansion of $(1+a)^{\mathrm{m}+\mathrm{n}}$, prove that the coefficients of $\mathrm{a}^{\mathrm{m}}$ and $\mathrm{a}^{\mathrm{n}}$ are equal.
Solution:
In the expansion of $(1+a)^{m+n}$.
Coefficient of $a^m={ }^{m+n} C_m=\frac{(m+n) !}{m !(m+n-m) !}=\frac{(m+n) !}{m ! n !}$
Coefficient of $a^n={ }^{m+n} C_n=\frac{(m+n) !}{n !(m+n-n) !}=\frac{(m+n) !}{n ! m !}$
$(1)=(2) \Rightarrow$ Coefficient of $a^m=$ Coefficient of $a^n$.
Question 14.
The coefficient of $(\mathrm{r}-1)^{\text {th }}, \mathrm{r}^{\text {th }}$, and $(\mathrm{r}+1)^{\text {th }}$ terms in the expansion of $(\mathrm{x}+1)^{\mathrm{n}}$ are in the ratio $1: 3: 5$. Find both $n$ and $r$.
Solution:
We know that co-effcients of $(\mathrm{r}-1)^{\text {th }}, \mathrm{r}^{\text {th }}$ and $(\mathrm{r}+1)^{\text {th }}$ terms in the expansion of $(\mathrm{x}+1)^{\mathrm{n}}$ are ${ }^{\mathrm{n}} C_{\mathrm{r}-2}$ : ${ }^{\mathrm{n}} C_{\mathrm{r}}$ -1 and ${ }^n \mathrm{C}_{\mathrm{r}}$ respectively
Now, ${ }^n \mathrm{C}_{r-2}:{ }^n \mathrm{C}_{r-1}:{ }^n \mathrm{C}_r=1: 3: 5$ (given)
$
\begin{aligned}
& \frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{5}{3} \text { and } \frac{{ }^n C_{r-1}}{{ }^n C_{r-2}}=\frac{3}{1} \\
\Rightarrow \quad & \frac{n-r+1}{r}=\frac{5}{3} \text { and } \frac{n-r+2}{r-1}=\frac{3}{1}
\end{aligned}
$

$
\begin{aligned}
& {\left[\therefore \frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}\right] } \\
& \Rightarrow 3 \mathrm{n}-8 \mathrm{r}+3=0 \text { and } \mathrm{n}-4 \mathrm{r}+5=0
\end{aligned}
$
Solving these for $\mathrm{n}, \mathrm{r}$ we get, $\mathrm{n}=7$ and $\mathrm{r}=3$
Question 5.
Find the value of $\left(a^2+\sqrt{a^2-1}\right)^4+\left(a^2-\sqrt{a^2-1}\right)^4$
Solution:
Suppose $a^2=x$ and $\sqrt{a^2-1}=y$
$
\therefore a^2+\sqrt{a^2-1}=x+y \text { and } a^2-\sqrt{a^2-1}=x-y
$
Using binomial theorem,

$
\begin{aligned}
& (x+y)^4={ }^4 \mathrm{C}_0 x^4+{ }^4 \mathrm{C}_1 x^3 y+{ }^4 \mathrm{C}_2 x^2 y^2+{ }^4 \mathrm{C}_3 x y^3+{ }^4 \mathrm{C}_4 y^4 \\
& \begin{aligned}
\Rightarrow(x+y)^4=x^4+4 x^3 y+6 x^2 y^2+4 x y^3+y^4
\end{aligned} \\
& \text { and }(x-y)^4={ }^4 \mathrm{C}_0 x^4-{ }^4 \mathrm{C}_1 x^3 y+{ }^4 \mathrm{C}_2 x^2 y^2-{ }^4 \mathrm{C}_3 x y^3+{ }^4 \mathrm{C}_4 y^4 \\
& \begin{aligned}
\Rightarrow(x-y)^4=x^4-4 x^3 y+6 x^2 y^2-4 x y^3+y^4
\end{aligned} \\
& \begin{aligned}
& \therefore(x+y)^4+(x-y)^4=2\left[x^4+6 x^2 y^2+y^4\right] \\
& \therefore\left(a^2+\sqrt{a^2-1}\right)^4+\left(a^2-\sqrt{a^2-1}\right)^4=2\left[\left(a^2\right)^4+6\left(a^2\right)^2\left(\sqrt{a^2-1}\right)^2+\left(\sqrt{a^2-1}\right)^4\right] \\
&=2\left[a^8+6 a^6-6 a^4+a^4-2 a^2+1\right]
\end{aligned} \\
& =2\left[a^8+6 a^6-5 a^4-2 a^2+1\right]
\end{aligned}
$
Question 6.
Show that the coefficient of the middle term in the expansion of $(1+x)^{2 n}$ is equal to the sum of the coefficients of the two middle terms in the expansion of $(1+x)^{2 n-1}$.
Solution:
In the expansion of $(1+x)^{2 n}$

Number of terms $=2 n+1$, which is odd
There is only one middle term, the $\frac{(2 n+1)+1}{2}$ th i.e. $(n+1)^{\text {th }}$ term
$\therefore T_{n+1}$ is the only middle term $T_{n+1}={ }^{2 n} C_n x^n$
Coefficient of $T_{n+1}={ }^{2 n} C_n$
In the expansion of $(1+x)^{2 n-1}$,
Number of terms $=2 n-1+1=2 n$, which is even,
There are two middle terms, the $\frac{2 n}{2}$ th i.e. $T_n$ and $T_{n+1}$
$T_n={ }^{2 n-1} C_{n-1} x^{n-1}$ and $T_{n+1}={ }^{2 n-1} C_n x^n$
Coefficient of $T_n={ }^{2 n-1} C_{n-1}$ and Coefficient of $T_{n+1}={ }^{2 n-1} C_n$
Sum of the coefficients of the two middle terms in the expansion of
$
\begin{aligned}
& (1+x)^{2 n-1}={ }^{2 n-1} C_n+{ }^{2 n-1} C_{n-1}={ }^{2 n} C_n \\
& \text { RHS of (1) = RHS of (2), }\left[\because{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right]
\end{aligned}
$
Question 7.
If three consecutive coefficients in the expansion of $(1+x)^{\mathrm{n}}$ are in the ratio $6: 33: 110$, find $n$. Solution:
Let the consecutive coefficients ${ }^n C_r,{ }^n C_r+1$ and ${ }^n C_r+2$ be the coefficients of $T_{r+1}, T_{r+2}$ and $T_{r+3}$ then ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}:{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}:{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+2}=6: 33: 110$

$
\begin{aligned}
& \text { Now, } \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{6}{33} \\
& \Rightarrow \frac{n !}{r !(n-r) !} \times \frac{(r+1) !(n-r-1) !}{n !}=\frac{2}{11} \\
& \Rightarrow \frac{r+1}{n-r}=\frac{2}{11} \\
& \Rightarrow 11 r+11=2 n-2 r \Rightarrow 2 n-13 r=11 \\
& \text { Again, } \frac{{ }^n C_{r+1}^n}{{ }^n C_{r+2}}=\frac{33}{110} \\
& \Rightarrow \frac{n !}{(r+1) !(n-r-1) !} \times \frac{(r+2) !(n-r-2) !}{n !}=\frac{3}{10} \\
& \Rightarrow \frac{(r+2)(r+1) !(n-r-2) !}{(r+1) !(n-r-1)(n-r-2) !}=\frac{3}{10} \\
& \Rightarrow \frac{r+2}{n-r-1}=\frac{3}{10} \\
& \Rightarrow 10(r+2)=3(n-r-1) \\
& \Rightarrow 3 n-13 r=23
\end{aligned}
$
Subtracting (2) from (1), we get $n=12$
Question 8 .
If the sum of the coefficients in the expansion of $(x+y)^{\mathrm{n}}$ is 4096 . Then find the greatest coefficient in the expansion.
Solution:
Given that, Sum of the coefficients in the expansion of $(x+y)^{\mathrm{n}}=4096$
$
\therefore{ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2+\ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}=4096
$
$\left[\therefore\right.$ Sum of binomial coefficients in the expansion of $(\mathrm{x}+\mathrm{a})^{\mathrm{n}}$ is $2^{\mathrm{n}}$ ]

$
\begin{aligned}
& \Rightarrow 2^{\mathrm{n}}=4096=2^{12} \\
& \Rightarrow n=12 \text { (even) }
\end{aligned}
$
So the greatest coefficient $=$ Coefficient of the middle term $\left(\frac{n}{2}+1\right)^{\text {th }}$ term
$=$ Coefficient of the middle term $\left(\frac{12}{2}+1\right)^{t h}$ term $=$ Coefficient of the $7^{\text {th }}$ term
Hence the greatest coefficient $={ }^{12} \mathrm{C}_6=\frac{(12) !}{6 !(12-6) !}=\frac{(12) !}{6 ! 6 !}=\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1}=924$
Question 9.
Find the coefficient of $\frac{1}{x^{17}}$ in the expansion of $\left(x^4-\frac{1}{x^3}\right)^{15}$.
Solution:
The general term in the expansion of
$
\mathrm{T}_{r+1}={ }^{15} \mathrm{C}_r\left(x^4\right)^{15-r}\left(-\frac{1}{x^3}\right)^r={ }^{15} \mathrm{C}_r(-1)^r x^{60-4 r} x^{-3 r}={ }^{15} \mathrm{C}_r(-1)^r x^{60-4 r-3 r}={ }^{15} C_r(-1)^r x^{60-7}
$
This will involve $x^{-17}$ if $60-7 r=-17 \Rightarrow 7 r=77 \Rightarrow r=11$
$
\therefore T_{11+1}={ }^{15} C_4(-1)^{11} x^{-17}=\frac{-15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} x^{-17}=-1365 x^{-17}
$
Hence, the coefficient of $\frac{1}{x^{17}}$ in the expansion of $\left(x^4-\frac{1}{x^3}\right)^{15}$ is -1365
Question 10.
If $p$ is a real number and if the middle term in the expansion of $\left(\frac{p}{2}+2\right)^8$ is 1120 , find $l$

Solution:
In the equation of $\left(\frac{p}{2}+2\right)^8$, Number of terms $=8+1=9$ (odd)
$\therefore$ There is only one middle term i.e. $\left(\frac{9+1}{2}\right)^{\text {th }}$ or $5^{\text {th }}$ term
$
\begin{aligned}
& T_{r+1}={ }^8 C_r\left(\frac{p}{2}\right)^r(2)^{8-r} \\
& \therefore T_5=T_{4+1}={ }^8 C_4\left(\frac{p}{2}\right)^4(2)^{8-4}=1120 \text { (Given) } \\
& \Rightarrow \frac{8 !}{4 !}\left(\frac{p}{2}\right)^4(2)^4=1120 ! \\
& \Rightarrow \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} p^4=1120 \Rightarrow 70 p^4=1120 \\
& \Rightarrow p^4=\frac{1120}{70}=16 \Rightarrow p^2=4, \text { so } p= \pm 2
\end{aligned}
$