Exercise 5.4 - Chapter 5 Binomial Theorem Sequences and Series 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 5.4
Question 1.
Expand the following in ascending powers of $\mathrm{x}$ and find the condition on $\mathrm{x}$ for which the binomial expansion is valid.
Solution:
(i) $\frac{1}{5+x}=\frac{1}{5\left(1+\frac{x}{5}\right)}=\frac{1}{5}\left(1+\frac{x}{5}\right)^{-1}=\frac{1}{5}\left\{1-\frac{x}{5}+\left(\frac{x}{5}\right)^2-\left(\frac{x}{5}\right)^3 \ldots\right\}$
Hence $\left|\frac{x}{5}\right|<1 \Rightarrow \therefore|x|<5$
$
=\frac{1}{5}-\frac{x}{5^2}+\frac{x^2}{5^3}-\frac{x^3}{5^4}
$
(ii) $\frac{2}{(3+4 x)^2}=\frac{2}{\left[3\left(1+\frac{4}{3} x\right)\right]^2}=\frac{2}{9\left(1+\frac{4}{3} x\right)^2}$
$
=\frac{2}{9}\left(1+\frac{4}{3} x\right)^{-2}=\frac{2}{9}\left[1-2\left(\frac{4}{3} x\right)+3\left(\frac{4}{3} x\right)^2 \cdots\right]
$
$
=\frac{2}{9}\left[1-\frac{8}{3} x+\frac{16}{9} x^2 \ldots\right]
$
Hence $\left|\frac{4 x}{3}\right|<1 \Rightarrow \therefore|x|<3 / 4$

(iii)
$
\begin{aligned}
\left(5+x^2\right)^{\frac{2}{3}} & =\left\{5\left(1+\frac{x^2}{5}\right)\right\}^{2 / 3} \\
& =5^{\frac{2}{3}}\left[\left(1+\frac{x^2}{5}\right)^{2 / 3}\right]=5^{2 / 3}\left\{1+\frac{2}{3}\left(\frac{x^2}{5}\right)+\frac{\frac{2}{3}\left(\frac{2}{3}-1\right)}{2.1}\left(\frac{x^2}{5}\right)^2 \cdots\right\}
\end{aligned}
$

$\begin{aligned}
& =5^{2 / 3}\left\{1+\frac{2 x^2}{15}-\frac{2}{9 \times 2}\left(\frac{x^4}{25}\right) \cdots\right\} \\
& =5^{\frac{2}{3}}\left\{1+\frac{2 x^2}{15}-\frac{x^4}{225} \ldots\right\} \\
& \text { Hence }\left|\frac{x^2}{5}\right|<1 \Rightarrow\left|x^2\right|<5
\end{aligned}$

(iv)

$
\begin{aligned}
(x+2)^{-\frac{2}{3}} & =\frac{1}{(x+2)^{2 / 3}}=\frac{1}{2^{2 / 3}\left(1+\frac{x}{2}\right)^{2 / 3}}=2^{-\frac{2}{3}}\left(1+\frac{x}{2}\right)^{-2 / 3} \\
& =2^{-2 / 3}\left(1-\frac{2}{3}\left(\frac{x}{2}\right)+\frac{\left(\frac{-2}{3}\right)\left(\frac{-2-1}{3}\right)\left(\frac{x}{2}\right)^2}{2 !}-\ldots\right) \\
& =2^{-2 / 3}\left(1-\frac{x}{3}+\frac{10}{18} \frac{x^2}{4}-\ldots .\right) \\
& =2^{-2 / 3}\left\{1-\frac{x}{3}+\frac{5 x^2}{36}-\frac{5}{81} x^3 \ldots\right\}
\end{aligned}
$
Hence $\left|\frac{x}{2}\right|<1 \Rightarrow|x|<2$
Question 2.
Find $\sqrt[3]{1001}$ approximately (two decimal places).
Solution:
$
\begin{aligned}
\sqrt[3]{1001} & =(1001)^{1 / 3}=(1000+1)^{1 / 3}=\left\{1000\left(1+\frac{1}{1000}\right)\right\}^{1 / 3}=(1000)^{1 / 3}\left[1+\frac{1}{10^3}\right]^{1 / 3} \\
& =10\left\{1+\frac{1}{3}\left(\frac{1}{10^3}\right)+\frac{\frac{1}{3}(-2 / 3)}{2}\left(\frac{1}{10^3}\right)^2 \cdots\right\} \\
& =10\left\{1+\frac{1}{3000}-\frac{2}{18000000} \ldots\right\}=10[1+0.000333 . .]=10(1.000333) \\
& =10.0033
\end{aligned}
$

Question 3.
Prove that $\sqrt[3]{x^3+6}-\sqrt[3]{x^3+3}$ is approximately equal to $\frac{1}{x^2}$ when $x$ is sufficiently large.
Solution:
When $x$ is large then $\frac{1}{x}$ will be small (i.e.) $\left|\frac{1}{x}\right|<1$
So $\sqrt[3]{x^3+6}-\sqrt[3]{x^3+3}=\left(x^3+6\right)^{1 / 3}-\left(x^3+3\right)^{1 / 3}$
$
\begin{aligned}
& =\left\{x^3\left(1+\frac{6}{x^3}\right)\right\}^{1 / 3}-\left\{x^3\left(1+\frac{3}{x^3}\right)\right\}^{1 / 3} \\
& =x\left(1+\frac{6}{x^3}\right)^{1 / 3}-x\left(1+\frac{3}{x^3}\right)^{1 / 3}=x\left[1+\frac{1}{3} \cdot \frac{6}{x^3} \ldots\right]-x\left[1+\frac{1}{3} \cdot \frac{3}{x^3} \ldots\right] \\
& =\left(x+\frac{2}{x^2} \ldots\right)-\left(x+\frac{1}{x^2} \ldots\right)=x+\frac{2}{x^2} \ldots-x-\frac{1}{x^2} \ldots \\
& =\frac{1}{x^2} \text { (approximately) }
\end{aligned}
$

Question 4.
Prove that $\sqrt{\frac{1-x}{1+x}}$ is approximately equal to $1-x+\frac{x^2}{2}$ when $x$ is very small.
Solution:
$
\begin{aligned}
& \text { LHS }=\sqrt{\frac{1-x}{1+x}}=\frac{(1-x)^{1 / 2}}{(1+x)^{1 / 2}}=(1-x)^{1 / 2}(1+x)^{-1 / 2} \\
& =\left(1-\frac{1}{2} x+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2 \cdot 1} x^2 \ldots \ldots .\right)\left(1-\frac{1}{2} x+\frac{\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)}{2 \cdot 1} x^2 \ldots .\right)=\left(1-\frac{x}{2}-\frac{x^2}{8} \ldots\right)\left(1-\frac{x}{2}+\frac{3 x^2}{8} \ldots\right) \\
& =1-\frac{x}{2}+\frac{3 x^2}{8}-\frac{x}{2}+\frac{x^2}{4}-\frac{x^2}{8} \ldots=\frac{1-2 x}{2}+\frac{4 x^2}{8}+\ldots \\
& =1-x+\frac{x^2}{2}=\mathrm{RHS}
\end{aligned}
$
Question 5.
Write the first 6 terms of the exponential series
(i) $\mathrm{e}^{5 \mathrm{x}}$
(ii) $e^{-2 x}$
(iii) $e^{\frac{x}{2}}$
Solution:

(i) $e^x=1+\frac{x}{\angle 1}+\frac{x^2}{\angle 2}+\frac{x^3}{\angle 3} \ldots$
So $e^{5 x}=1+\frac{5 x}{\angle 1}+\frac{(5 x)^2}{\angle 2}+\frac{(5 x)^3}{\angle 3}+\frac{(5 x)^4}{\angle 4}+\ldots$.
$
\begin{aligned}
& =1+5 x+\frac{25 x^2}{2}+\frac{125 x^3}{6}+\frac{625 x^4}{24}+\frac{3125}{120} x^5+\frac{15625}{720} x^6 \ldots \\
& =1+5 x+\frac{25 x^2}{2}+\frac{125 x^3}{6}+\frac{625 x^4}{24}+\frac{625 x^5}{24}+\frac{3125 x^6}{144} \ldots
\end{aligned}
$
(ii)
$
\begin{aligned}
e^{-2 x} & =1+\frac{(-2 x)}{\angle 1}+\frac{(-2 x)^2}{\angle 2}+\frac{(-2 x)^3}{\angle 3}-\ldots \\
& =1-2 x+\frac{4 x^2}{2}-\frac{8 x^3}{6}+\frac{16 x^4}{24}-\frac{32 x^5}{120}+\frac{64 x^6}{720}-\ldots \\
& =1-2 x+2 x^2-\frac{4 x^3}{3}+\frac{2 x^4}{3}-\frac{4 x^5}{15}+\frac{4 x^6}{45}-\ldots
\end{aligned}
$
(iii)
$
\begin{aligned}
e^{\frac{1}{2} x} & =1+\frac{1}{2} x+\frac{\left(\frac{1}{2} x\right)^2}{\angle 2}+\frac{\left(\frac{1}{2} x\right)^3}{\angle 3} \ldots \\
& =1+\frac{x}{2}+\frac{x^2}{8}+\frac{x^3}{48}+\frac{x^4}{384}+\frac{x^5}{3840}+\ldots
\end{aligned}
$

Question 6 .
Write the first 4 terms of the logarithmic series
(i) $\log (1+4 x)$
(ii) $\log (1-2 x)$,
(iii) $\log \left(\frac{1+3 x}{1-3 x}\right)$
(iv) $\log \left(\frac{1-2 x}{1+2 x}\right)$.
Find the intervals on which the expansions are valid.
Solution:

(i) $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} \ldots, \log (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots$
$
\begin{aligned}
\log (1+4 x) & =4 x-\frac{(4 x)^2}{2}+\frac{(4 x)^3}{3}-\frac{(4 x)^4}{4} \ldots \\
\text { Hence }|4 x| & <1 \Rightarrow|x|<1 / 4 \\
& =4 x-\frac{16 x^2}{2}+\frac{64 x^3}{3}-\frac{256 x^4}{4} \ldots \\
& =4 x-8 x^2+\frac{64}{3} x^3-64 x^4 \ldots
\end{aligned}
$

$
\text { (ii) } \begin{aligned}
\log (1-2 x) & =-2 x-\frac{(2 x)^2}{2}-\frac{(2 x)^3}{3}-\frac{(2 x)^4}{4} \ldots=-2 x-\frac{4 x^2}{2}-\frac{8 x^3}{3}-\frac{16 x^4}{4} \ldots \\
& =-2 x-2 x^2-\frac{8 x^3}{3}-4 x^4
\end{aligned}
$
Hence $|2 x|<1 \Rightarrow|x|<1 / 2$

$
\text { (iii) } \begin{aligned}
\log \left(\frac{1+3 x}{1-3 x}\right)= & \log (1+3 x)-\log (1-3 x) \\
& =\left[3 x-\frac{(3 x)^2}{2}+\frac{(3 x)^3}{3}-\frac{(3 x)^4}{4} \ldots .\right]-\left[-\frac{(3 x)^2}{2}-\frac{(3 x)^3}{3}-\frac{(3 x)^4}{4} \ldots\right] \\
& =3 x-\frac{(3 x)^2}{2}+\frac{(3 x)^3}{3}-\frac{(3 x)^4}{4} \ldots+3 x+\frac{(3 x)^2}{2}+\frac{(3 x)^3}{3}+\ldots \\
& =2\left(3 x+\frac{(3 x)^3}{3}+\frac{(3 x)^5}{5}+\frac{(3 x)^7}{7} \ldots\right)
\end{aligned}
$
Hence $|3 x|<1 \Rightarrow|x|<1 / 3$

(iv)
$
\begin{aligned}
\log \left(\frac{1-2 x}{1+2 x}\right) & =\log (1-2 x)-\log (1+2 x) \\
& =\left[-2 x-\frac{(2 x)^2}{2}-\frac{(2 x)^3}{3}-\frac{(2 x)^4}{4} \ldots\right]-\left[2 x-\frac{(2 x)^2}{2}+\frac{(2 x)^3}{3}-\frac{(2 x)^4}{4} \ldots\right] \\
& =-2 x-\frac{(2 x)^2}{2}-\frac{(2 x)^3}{3}-\frac{(2 x)^4}{4} \ldots .-2 x+\frac{(2 x)^2}{2}-\frac{(2 x)^3}{3}+\frac{(2 x)^4}{4} \ldots \\
& =-2\left(2 x+\frac{(2 x)^3}{3}+\frac{(2 x)^5}{5}+\frac{(2 x)^7}{7} \ldots\right)
\end{aligned}
$
Hence $|2 x|<1 \Rightarrow|x|<1 / 2$

Question 7.
If $y=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\ldots$ then show that $x=y-\frac{y^2}{2 !}+\frac{y^3}{3 !}-\frac{y^4}{4 !}+\ldots$
Solution:
$
\begin{aligned}
& y=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\ldots \\
& \text { (i.e) } y=-\left[-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4} \ldots\right]=-\log (1-x)
\end{aligned}
$
(i.e) $y=-\log (1-x)=\log \frac{1}{1-x}$
So $\log \frac{1}{1-x}=y$
$
\Rightarrow \frac{1}{1-x}=e^y \Rightarrow 1-x=\frac{1}{e^y}=e^{-y} \Rightarrow 1-x=e^{-y} \Rightarrow 1-e^{-y}=x
$
(i.e) $x=1-\left[1-y+\frac{y^2}{2 !}-\frac{y^3}{3 !}+\frac{y^4}{4 !} \ldots\right]=1-1+y-\frac{y^2}{2 !}+\frac{y^3}{3 !}-\frac{y^4}{4 !} \ldots$
(i.e) $x=y-\frac{y^2}{2 !}+\frac{y^3}{3 !}-\frac{y^4}{4 !} \ldots$

Question 8.
If $\boldsymbol{p}-\boldsymbol{q}$ is small compared to either $\boldsymbol{p}$ or $\boldsymbol{q}$, then show that $\sqrt[n]{\frac{p}{q}} \simeq \frac{(n+1) p+(n-1) q}{(n-1) p+(n+1) q}$. Hence find $\sqrt[8]{\frac{15}{16}}$
Solution:
$
\begin{aligned}
\text { RHS } & =\frac{(n+1) p+(n-1) q}{(n-1) p+(n+1) q}=\frac{n(p+q)+(p-q)}{n(p+q)-(p-q)}=\frac{1+\frac{1}{n}\left(\frac{p-q}{p+q}\right)}{1-\frac{1}{n}\left(\frac{p-q}{p+q}\right)}=\frac{\left(1+\frac{p-q}{p+q}\right)^{\prime}}{\left(1-\frac{p-q}{p+q}\right)^n} \\
& =\left(\frac{p}{q}\right)^{1 / n}=\sqrt[n]{\frac{p}{q}}=\text { LHS }
\end{aligned}
$
To find $\sqrt[8]{\frac{15}{16}}$ we take $n=8, p=15, q=16$
So $\sqrt[8]{\frac{15}{16}}=\frac{(n+1) p+(n-1) q}{(n-1) p+(n+1) q}=\frac{9 \times 15+7 \times 16}{7 \times 15+9 \times 16}=\frac{135+112}{105+144}=\frac{247}{249}=0.99196$
Question 9.
Find the coefficient of $x^4$ in the expansion of $\frac{3-4 x+x^2}{e^{2 x}}$.

Solution:
$
\begin{aligned}
& \frac{3-4 x+x^2}{e^{2 x}}=\left(3-4 x+x^2\right) e^{-2 x} \\
&=\left(3-4 x+x^2\right)\left[1+\frac{(-2 x)}{1 !}+\frac{(-2 x)^2}{\angle 2}+\frac{(-2 x)^3}{43} \ldots\right] \\
& \text { Coefficient of } x^4: 3\left[\frac{(-2)^4}{4 !}\right]-4\left[\frac{(-2)^3}{3 !}\right]+1\left(\frac{(-2)^2}{2 !}\right) \\
&=3\left[\frac{16}{24}\right]+(-4) \frac{(-8)}{6}+\frac{4}{2}=\frac{48}{24}+\frac{32}{6}+2 \\
&=2+\frac{16}{3}+2=\frac{6+16+6}{3}=\frac{28}{3}
\end{aligned}
$
Question 10.
Find the value of $\sum_{n=1}^{\infty} \frac{1}{2 n-1}\left(\frac{1}{9^{n-1}}+\frac{1}{9^{2 n-1}}\right)$.
Solution:

$
\begin{aligned}
S_{\infty} & =\frac{1}{1}\left(1+\frac{1}{9}\right)+\frac{1}{3}\left(\frac{1}{9}+\frac{1}{9^2}\right)+\frac{1}{5}\left(\frac{1}{9^2}+\frac{1}{9^5}\right)+\frac{1}{7}\left(\frac{1}{9^3}+\frac{1}{9^7}\right)+\ldots \\
& =\left(1+\frac{1}{3} \cdot \frac{1}{9}+\frac{1}{5} \cdot \frac{1}{9^2}+\ldots\right)+\left(\frac{1}{9}+\frac{1}{3} \cdot \frac{1}{9^3}+\frac{1}{5} \cdot \frac{1}{9^5}\right) \cdots \\
& =\left(1+\frac{1}{3} \cdot \frac{1}{3^2}+\frac{1}{5} \cdot \frac{1}{3^4}+\frac{1}{7} \cdot \frac{1}{3^6}+\ldots\right)+\left[\frac{1}{9}+\frac{1}{3}\left(\frac{1}{9}\right)^3+\frac{1}{5}\left(\frac{1}{9}\right)^5+\ldots\right]
\end{aligned}
$
Multiplying and dividing the I summation by 3 we get
$
\begin{aligned}
& 3\left(\frac{1}{3}+\frac{1}{3} \cdot \frac{1}{3^3}+\frac{1}{5} \cdot \frac{1}{3^5}+\ldots\right)+\frac{1}{9}+\frac{1}{3}\left(\frac{1}{9}\right)^3+\ldots \\
& =\left[3\left(\frac{1}{3}\right)+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\ldots+\left[\frac{1}{9}+\frac{(1 / 9)^3}{3}+\frac{(1 / 9)^5}{5}+\ldots\right]\right. \\
& \left\{\text { We know } 1 / 2 \log \left(\frac{1+x}{1-x}\right)=x+\frac{x^3}{3}+\frac{x^5}{5}+\ldots\right\} \\
& =3 \times \frac{1}{2} \log \left(\frac{1+1 / 3}{1-1 / 3}\right)+\frac{1}{2} \log \left(\frac{1+1 / 9}{1-1 / 9}\right) \\
& =\frac{1}{2}\left\{3 \log \frac{4 / 3}{2 / 3}\right\}+\frac{1}{2} \log \frac{10}{8 / 9}=\frac{1}{2}\left[3 \log 2+\log \frac{10}{8}\right]=\frac{1}{2}\left[\log 2^3+\log \frac{10}{8}\right] \\
& =\frac{1}{2}\left[\log 8 \cdot \frac{10}{8}\right]=\frac{1}{2} \log e^{10}[\log a+\log b=\log a b]
\end{aligned}
$