# Question 1 Exercise 5.3

Solutions of Question 1 of Exercise 5.4 of Unit 05: Mascellaneous series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find the sum of the series $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\ldots$ to $n$ terms.

The general term of the series is: $$T_n=\dfrac{1}{n(n+1)}$$ Resolving $T_n$ into partial fractions $$\dfrac{1}{n(n+1)}=\dfrac{A}{n}+\dfrac{B}{(n+1)}$$ Multiplying both sides by $n(n+1)$ $$1=A(n+1)+B n=(A+B) n+A$$ Comparing the coeffioients of $n$ and constants on the both sides of the above equation, we get $$A+B=0 \text{and} A=1$$ Putting $A=1$,then \begin{align}1+B&=0\\ B&=-1\end{align} hence \begin{align} & \dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1} \\ & \Rightarrow T_n=\dfrac{1}{n}-\dfrac{1}{n+1}\end{align} Taking summation of the both sides of the above equation \begin{align} \sum_{k=1}^n T_k&=\sum_{k=1}^n(\dfrac{1}{k}-\dfrac{1}{k+1}) \\ & =(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+(\dfrac{1}{3}-\dfrac{1}{4}) \\ & +\ldots+(\dfrac{1}{n}-\dfrac{1}{n+1}) \\ & =1-\dfrac{1}{n+1} \\ & =\dfrac{n}{n+1} \end{align} Hence the sum of the series is: $$S_n=\dfrac{n}{n+1}$$

Find the sum of the series $\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\ldots$ to $n$ terms.

Here $n$ term of the series is: $u_n=\dfrac{1}{(2 n-1)(2 n+1)}$.

Resolving into partial fractions $$\dfrac{1}{(2 n-1)(2 n-1)}=\dfrac{A}{(2 n-1)}+\dfrac{B}{(2 n+1)}$$ Multiplying both sides by $(2 n-1)(2 n+1)$. we get, \begin{align} & \mathrm{I}=A(2 n+1)+B(2 n-1) \\ & =(2 A+2 B) n+A-B\end{align} Comparing coefficients of $n$ and constant term $$2 A+2 B=0 \text { and } A-B=1$$ Solving the above two equations for $A$ and $B$, we get \begin{align}A&=\dfrac{1}{2}\\ \text{and} B&=-\dfrac{1}{2}\end{align} Hence $$u_n=\dfrac{1}{2}[\dfrac{1}{(2 n-1)}-\dfrac{1}{(2 n+1)}]$$ Taking summation on the both sides \begin{align} S_n&=\dfrac{1}{2} \sum_{r=1}^n[\dfrac{1}{(2 r-1)}-\dfrac{1}{(2 r+1)}] \\ & =\dfrac{1}{2}[(1-\dfrac{1}{3})+(\dfrac{1}{3}-\dfrac{1}{5}) \\ & +(\dfrac{1}{5}-\dfrac{1}{7})+\ldots+(\dfrac{1}{(2 n-1)}-\dfrac{1}{(2 n+1)})] \\ & =\dfrac{1}{2}[1 \cdots \dfrac{1}{(2 n+1)}] \\ & =\dfrac{1}{2}(\dfrac{2 n+1-1}{2 n+1}) \\ \Rightarrow S_n&=\dfrac{n}{2 n+1}\end{align}

Find the sum of the series $\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\ldots$ to infinity.

Here in the denominator the factors are the product of two successive terms of the A.P 2,5,8,11,…

Therefore, the general term of the series is: $$u_n=\dfrac{1}{(3 n-1)(3 n+2)}$$ Resolving into partial fractions: $$\dfrac{1}{(3 n-1)(3 n+2)}=\dfrac{A}{3 n-1}-\dfrac{B}{3 n+2}$$ Multiplying both sides by $(3 n-1)(3 n+2)$ we get, \begin{align} 1&=A(3 n+2)+B(3 n-1) \\ \Rightarrow(3 A+3 B) n+2 A-B&=1\end{align} Comparing the coefficients of $n$ and constants on the both sides of the above equation,

we get $$3 A+3 B=0 \quad \text{and} \quad 2 A-B=1$$ Solving the above two equations $$A=\dfrac{1}{3}\quad\text{and} B=-\dfrac{1}{3}$$ $$u_n=\dfrac{1}{3}[\dfrac{1}{3 n-1}-\dfrac{1}{3 n+2}]$$ Taking summation of the both sides \begin{align}\sum_{r=1}^n u_n&=\dfrac{1}{3} \sum_{r=1}^n[\dfrac{1}{3 r-1}-\dfrac{1}{3 r+2}] \\ & =\dfrac{1}{3}[(\dfrac{1}{2}-\dfrac{1}{5})+(\dfrac{1}{5}-\dfrac{1}{7}) \\ & +\cdots+(\dfrac{1}{3 n-1}-\dfrac{1}{3 n+2})] \\ & =\dfrac{1}{3}[\dfrac{1}{2}-\dfrac{1}{3 n-2}]\\ S_n&=\dfrac{1}{3}[\dfrac{1}{2}-\dfrac{1}{3 n+2}]\end{align}

Taking limit $n \longrightarrow \infty$, we get $$S_{\infty}=\dfrac{1}{6}$$

Find the sum of the series $\dfrac{1}{4.13}+\dfrac{1}{13.22}+\dfrac{1}{22.31}+\ldots$ to infinity.

Here in the deneminator the factors are the product of two successive terms of the A.P $4.13,22, \ldots$.

Therefore, the general term of the series is: $u_n=\dfrac{1}{(9 n-5)(9 n+4)}$

Resolving into partial fractions: $$\dfrac{1}{(9 n-5)(9 n+4)}=\dfrac{A}{9 n-5}+\dfrac{B}{9 n+4}$$ Multiplying both sides by $(9 n-5)(9 n+4)$ we get \begin{align} 1&=A(9 n+4)+B(9 n-5) \\ \Rightarrow(9 A+9 B) n+4 A-5 B&=1\end{align} Comparing the cocficients of $n$ and constants on the both sides of the above equation, we get

$$9 A+9 B=0\quad \text{ and} \quad 4 A-5 B=1$$ Solving the above equations for $A$ and $B$

we get $A=\dfrac{1}{9}$ and $B=-\dfrac{1}{9}$.

Thus $$u_n=\dfrac{1}{9}[\dfrac{1}{9 n-5}-\dfrac{1}{9 n+4}]$$ Taking summation of the both sides \begin{align} \sum_{r=1}^n u_n&=\dfrac{1}{9} \sum_{r=1}^n[\dfrac{1}{9 n-5}-\dfrac{1}{9 n+4}] \\ & =\dfrac{1}{9}[(\dfrac{1}{4}-\dfrac{1}{13})+(\dfrac{1}{13}-\dfrac{1}{22})+(\dfrac{1}{22}-\dfrac{1}{31}) \\ & +\cdots+(\dfrac{1}{9 n-5}-\dfrac{1}{9 n+4})] \\ & =\dfrac{1}{9}[\dfrac{1}{4}-\dfrac{1}{9 n+4}]\\ S_n&=\dfrac{1}{9}[\dfrac{1}{4}-\dfrac{1}{9 n+13}]\end{align}

Taking limit $n \longrightarrow \infty$ both sides $$S_{\infty}=\dfrac{1}{9} \cdot \dfrac{1}{4}=\dfrac{1}{36}$$