Exercise 16.3
Question 1
(i) \frac{e^{2 x}+1}{2 e^{x}}
Sol :
=\frac{1}{2}\left[\frac{e^{2 x}+1}{e^{x}}\right]
=\frac{1}{2}\left[\frac{e^{2 x}}{e^{x}}+\frac{1}{e^{2}}\right]
=\frac{1}{2}\left[e^{x}+e^{-x}\right]
=\frac{1}{2}\left[2 \cdot\left\{1+\frac{x^{2}}{2!}+\frac{x^{4}}{4 !}+\cdots
\cdot \infty\right\}\right]
=1+\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\cdots \cdot \infty
(ii) \frac{e^{x}-e^{-x}}{2}
Sol :
=\frac{e^{x}\left(e^{4 x}+1\right)}{e^{3 x}}-1
=\frac{e^{4 x}+1}{e^{2 x}}-1
=\frac{e^{4 x}}{e^{2 z}}+\frac{1}{e^{2 x}}-1
=e^{2 x}+e^{-2 x}-1
=2\left[1+\frac{(2 x)^{2}}{2 !}+\frac{(2x)^{4}}{4 !}+\cdots \cdot
\infty\right]-1
=2\left[1+\frac{4x^{2}}{2 !}+\frac{16 x^{4}}{4 !}+\dots \infty\right]-1
=2+\frac{8 x^{2}}{2 !}+\frac{32 x^{4}}{4 !}+\dots \infty-1
=1+\frac{8 a^{2}}{2!}+\frac{32 x^{4}}{4 !}+\cdots \dots \infty
(iii) \frac{e^{5 x}+e^{x}}{e^{3 x}}-1
Sol :
(iv) \frac{e^{7 x}+e^{x}}{e^{4 x}}
Sol :
Question 2
\sum_{n=1}^{\infty} \underline{n} का मान निकाले
[Find the value of \sum_{n=1}^{\infty} \underline{n}]
Sol :
\sum_{n=1}^{\infty} \frac{n^{2}}{n !}=\frac{1^{2}}{1 !}+\frac{2^{2}}{2
!}+\frac{3^{2}}{3 !}+\dots \infty
T_{n}=\frac{n^{2}}{n !}=\frac{n^{2}}{n(n-1) !}
T_{n}=\frac{n}{(n-1) !}=\frac{(n-1)+1}{(n-1) !}
T_{n}=\frac{n-1}{(n-1) !}+\frac{1}{(n-1) !}
T_{n}=\frac{n-1}{(n-1)(n-2) !}+\frac{1}{(n-1) !}
T_{n}=\frac{1}{(n-2) !}+\frac{1}{(n-1) !}
\begin{aligned}T_{1}&=0+\frac{1}{0 !}\\T_{2}&=\frac{1}{0
!}+\frac{1}{1 !}\\ \cdot &\quad \cdot \quad \cdot \\ \cdot &\quad
\cdot \quad \cdot \\ \cdot &\quad \cdot \quad \cdot \\ \infty
&=\cdot \quad \cdot \\ \hline \sum_{n=1}^{\infty} \frac{n^{2}}{n
!}&=e+e=2 e \end{aligned}
Question 3
साबित कीजिए कि (prove that) \frac{e^{x}-1}{x}=1+\frac{x}{2
!}+\frac{x^{2}}{3 \cdot !}+\ldots ~to~\infty$
Sol :
L.H.S
\frac{e^{x}-1}{x}=\frac{1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3
!}+\cdots \cdot \infty-1}{x}
=\frac{x\left(1+\frac{x}{2 !}+\frac{x^{2}}{3 !}+\cdots \cdot
\infty\right)}{x}
=1+\frac{x}{2 !}+\frac{x^{2}}{3 !}+ \dots \infty
Question 4
Question 5
साबित कीजिए कि (Prove that)
\frac{a^{x}-1}{x}=\log a+\frac{x(\log a)^{2}}{2 !}+\ldots to \infty
Sol :
L.H.S
\frac{a^{x}-1}{x}=\frac{1+\frac{x \log a}{11}+\frac{x^{2}(\log a)^{2}}{2
!}+\cdots\infty -1}{x}
=x\left(\frac{\log a+\frac{x(\log a)^{2}}{2
!}+\ldots~to~\infty}{\frac{x}{x}}\right)
=\log a+\frac{x(\log a)^{2}}{2!}+\dots ~\infty
Question 6
निम्नलिखित को सिद्ध कीजिए
(Prove the following ):
(i) \frac{1+\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots \infty \text
{ पद तक }}{\frac{1}{1 !}+\frac{1}{3 !}+\frac{1}{5 !}+\frac{1}{7 !}+\ldots
\infty \text { पद तक }}=\frac{e^{2}+1}{e^{2}-1}
Sol :
L.H.S
\frac{1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6 !}+\dots \infty}{\frac{1}{1
!}+\frac{1}{3 !}+\frac{1}{5 !}+\frac{1}{6!}\dots \infty}
=\frac{e+e^{-1}}{e-e^{-1}}=\frac{e+\frac{1}{e}}{e-\frac{1}{e}}
=\frac{\frac{e^{2}+1}{e}}{\frac{e^{2}-1}{e}}
=\frac{e^{2}+1}{e^{2}-1}
Question 7
Find the sum of the following series
(i) \frac{1}{3 !}+\frac{2}{5 !}+\frac{3}{7 !}+\frac{4}{9 !}+\ldots to
\infty
Sol :
दी गई श्रेणी का n वाँ पद
t_{n}=\frac{n}{(2 n+1) !}=\frac{1}{2} \cdot \frac{2 n}{(2 n+1) !}
=\frac{1}{2}\left[\frac{(2 n+1)-1}{(2 n+1) !}\right]
t_{n}=\frac{1}{2}\left[\frac{2 n+1}{(2 n+1)!}-\frac{1}{(2 n+1) !}\right]
t_{n}=\frac{1}{2}\left[\frac{2 n+{1}}{(2 n+1)(2 n) !}-\frac{1}{(2 n+1)
!}\right]
t_{n}=\frac{1}{2}\left[\frac{1}{(2 n) !}-\frac{1}{(2 n+1) !}\right]
\begin{aligned}t_{1}&=\frac{1}{2}\left[\frac{1}{2 !}-\frac{1}{3
!}\right] \\t_{2}&=\frac{1}{2}\left[\frac{1}{4 !}-\frac{1}{5
!}\right]\\t_{3}&=\frac{1}{2}\left[\frac{1}{6 !}-\frac{1}{7 !}\right]\\
\cdot &\quad \cdot \quad \cdot \\ \cdot &\quad \cdot \quad \cdot
\cdot &\quad \cdot \quad \cdot \\ \hline \end{aligned}
S_{n}=\frac{1}{2}\left[\left(\frac{1}{2 !}+\frac{1}{4!}+\frac{1}{6
!}+\cdots \infty\right)-\left(\frac{1}{3 !}+\frac{1}{5
!}+\frac{1}{7!}+\ldots \infty \right)\right]
S_{n}=\frac{1}{2}\left[\left(\frac{e+e^{-1}}{2}-1\right)-\left(\frac{e-e^{-1}}{2}-1\right)\right]
=\frac{1}{2}\left[\frac{e+e^{-1}}{2}-1-\frac{e-e^{-1}}{2}+1\right]
=\frac{1}{2}\left[\frac{e+e^{-1}-e+e^{-1}}{2}\right]
=\frac{1}{2}\left(\frac{2e^{-1}}{2}\right)=\frac{1}{2e}
(ii) \frac{1}{2 !}+\frac{3}{4 !}+\frac{5}{6 !}+\frac{7}{8 !}+\ldots to
\infty
Sol :
t_{n}=\frac{n}{(n+1) !}=\frac{(n+1)-1}{(n+1)!}
t_{n}=\left[\frac{n+1}{(n+1)!}-\frac{1}{(n+1) !}\right]
(iii) \frac{1}{2 !}+\frac{1+2}{3 !}+\frac{1+2+3}{4 !}+\ldots to
\infty
Sol :
दी गई श्रेणी का n वाँ पद
t_{n}=\frac{1+2+3+\cdots+n}{(n+1) !}
=\frac{n(n+1)}{2(n+1) !}
t_{n}=\frac{n(n+1)}{2(n+1)n \cdot(n-1) !}
t_{n}=\frac{1}{2(n-1) !}
$\begin{aligned}t_{1}=\frac{1}{2 \cdot 0 !}=\frac{1}{2} \cdot 1\\
t_{2}=\frac{1}{2.1 !}=\frac{1}{2} \cdot \frac{1}{1!}\\ t_{3}=\frac{1}{2.2
!}=\frac{1}{2} \cdot \frac{1}{2!}\\ t_{4}=\frac{1}{2 \cdot 3 !}=\frac{1}{2}
\cdot \frac{1}{3 !} \\ \cdot &\quad \cdot \quad \cdot \\ \hline
S_{n}=\frac{1}{2}\left[1+\frac{1}{1!}+\frac{1}{2 !}+\frac{1}{3 !}+\dots
\infty\right]\end{aligned}
=\frac{1}{2}(e)=\frac{e}{2}
(iv) \frac{1^{3}}{1 !}+\frac{2^{3}}{2 !}+\frac{3^{3}}{3 !}+\ldots
\infty
Sol :
दी गई श्रेणी का n वाँ पद
t_{n}=\frac{n^{3}}{n !}=\frac{n^{3}}{n(n-1) !}=\frac{n^{2}}{(n-1) !}
t_{n}=\frac{n \cdot n}{(n-1) !}
t_{n}=\frac{n[(n-1)+1]}{(n-1) !}
t_{n}=\frac{n(n-1)+n}{(n-1) !}
t_{n}=\frac{n(n-1)}{(n-1) !}+\frac{n}{(n-1)!}
t_{n}=\frac{n(n-1)}{(n-1)(n-2) !}+\frac{n}{(n -1)!}
t_{n}=\frac{(n-2)+2}{(n-2) !}+\frac{(n-1)+1}{(n-1) !}
t_{n}=\frac{n-2}{(n-2)(n-3) !}+\frac{2}{(n-2) !}+\frac{n-1}{(n-1)(n-2)
!}+\frac{1}{(n-1)!}
t_{n}=\frac{1}{(n-3) !}+\frac{2}{(n-2)}!+\frac{1}{(n-2) !}+\frac{1}{(n-1) !}
t_{n}=\frac{1}{(n-3) !}+\frac{3}{(n-2) !}+\frac{1}{(n-1) !}
t1=0+0+1
t2=0+3+\frac{1}{1!}
t_{3}=1+\frac{3}{1 !}+\frac{1}{2 !}
t_{4}=\frac{1}{1 !}+\frac{3}{2 !}+\frac{1}{3 !}
t_{5}=\frac{1}{2 !}+\frac{3}{3 !}+\frac{1}{4 !}
.
.
.
∞
\overline{S_{n}=\left[1+\frac{1}{1!}+\frac{1}{2!}+\dots\infty\right]+3\left[1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3!}+\dots \infty\right] +\left[1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\infty\right]}
=e+3e+e
=5e
(v) 1+\frac{1+3}{2 !}+\frac{1+3+3^{2}}{3 !}+\ldots to \infty
Sol :
दी गई श्रेणी का n वाँ पद
t_{n}=\frac{1+3+3^{2}+\cdots-3^{n}}{n !}
t_{n}=\frac{1\left(3^{n}-1\right)}{(3-1) \cdot n !}
t_{n}=\frac{1}{2} \cdot \frac{3^{n}-1}{n !}
t_{n}=\frac{1}{2}\left[\frac{3^{n}}{n{!}}-\frac{1}{n !}\right]
t_{1}=\frac{1}{2}\left[\frac{3^{1}}{1 !}-\frac{1}{1!}\right]
t_2=\frac{1}{2}\left[\frac{3^{2}}{2 !}-\frac{1}{2!}\right]
t_{3}=\frac{1}{2}\left[\frac{3^{3}}{3!}-\frac{1}{3 !}\right]
.
.
.
∞
\overline{S_{n}=\frac{1}{2}\left[\left(\frac{3^{1}}{1 !}+\frac{3^{2}}{2 !}+\frac{3^{3}}{3 !}+\cdots \infty\right)-\left(\frac{1}{1 !}+\frac{1}{2!}+\frac{1}{3!}+\dots \infty\right)\right.}
S_{n}=\frac{1}{2}\left[\left(e^{3}-1\right)-(e-1)\right]
=\frac{1}{2}\left[e^{3}-1-e+1\right]
=e \frac{\left(e^{2}-1\right)}{2}
(x) \frac{1 \cdot 2}{1 !}+\frac{2 \cdot 3}{2 !}+\frac{3 \cdot 4}{3 !}+\frac{4 \cdot 5}{4 !}+\ldots to \infty
Sol :
दी गई श्रेणी का n वाँ पद
t_{n}=\frac{n(n+1)}{n !}=\frac{ n(n+1)}{n(n-1)!}
t_{n}=\frac{n-1+2}{(n-1) !}
t_{n}=\frac{n-1}{(n-1) !}+\frac{2}{(n-1) !}
t_{n}=\frac{n-1}{(n-1)(n-2) !}+\frac{2}{(n-1) !}
t_{n}=\frac{1}{(n-2) !}+\frac{2}{(n-1) !}
t1=0+2
t_{2}=1+\frac{2}{1 !}
t_{3}=\frac{1}{1 !}+\frac{2}{2 !}
t_{4}=\frac{1}{2!}+\frac{2}{3 !}
.
.
.
∞
\overline{S_{n}=\left[1+\frac{1}{1!}+\frac{1}{2 !}+\ldots \infty\right]+2\left[1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\dots \infty\right]}
=e+2e=3e
(xii) \frac{1^{2}}{2 !}+\frac{1^{2}+2^{2}}{3 !}+\frac{1^{2}+2^{2}+3^{2}}{4 !}+\ldots to \infty
Sol :
दी गई श्रेणी का n वाँ पद
t_{n}=\frac{1^{2}+2^{2}+3^{2}+\cdots+n^{2}}{(n+1) !}
t_{n}=\frac{n(n+1)(2 n+1)}{6(n+1) !}
t_{n}=\frac{n(n+1)(2 n+1)}{6 \cdot(n+1) \cdot n(n-1) !}
t_{n}=\frac{(2 n+1)}{6(n-1) !}=\frac{2 n-2+3}{6(n-1) !}
t_{n}=\frac{2(n-1)+3}{6(n-1) !}
t_{n}=\frac{2(n-1)}{6(n-1) !}+\frac{3}{6(n-1) !}
t_{n}=\frac{n-1}{3(n-1)(n-2) !}+\frac{1}{2(n-1) !}
t_{n}=\frac{1}{3} \cdot \frac{1}{(n-2) !}+\frac{1}{2} \cdot \frac{1}{(n-1) !}
t_{1}=\frac{1}{3} \cdot 0+\frac{1}{2} \cdot 1
t_{2}=\frac{1}{3} \cdot 1 +\frac{1}{2} \cdot \frac{1}{1 !}
t_{3}=\frac{1}{3} \cdot \frac{1}{1 !}+\frac{1}{2} \cdot \frac{2}{2 !}
.
.
.
∞
\overfline{S_{n}=\frac{1}{3}\left[0+1+\frac{1}{1 !}+\frac{1}{2 !}\dots\infty\right]+\frac{1}{2}}\left[1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots \infty \right]
=\frac{1}{3}(e)+\frac{1}{2}(e)=\frac{e}{3}+\frac{e}{2}
=\frac{2 e+3 e}{6}=\frac{5 e}{6}
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