# Question 2 & 3 Exercise 5.4

Solutions of Question 2 & 3 of Exercise 5.4 of Unit 05: Miscullaneous 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 sum of the series: $\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2}$

\begin{align}\text { Let } S_n&=\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2} \\ S_n&=\sum_{k=1}^n \dfrac{1}{9 k^2+6 k-3 k-2} \\ & =\sum_{k=1}^n \dfrac{1}{3 k(3 k+2)-1(3 k+2)} \\ S_n&=\sum_{k=1}^n \dfrac{1}{(3 k-1)(3 k-2)} \end{align} The $n$ term of the above series is: $$u_n=\dfrac{1}{(3 k-1)(3 k+2)}$$ Resolving into partial fractions $$\dfrac{1}{(3 k-1)(3 k+2)}=\dfrac{A}{3 k-1}+\dfrac{B}{3 k+2}$$ Multiplying both sides by \begin{align} & (3 k-1)(3 k+2) \text { we get } \\ & 1=A(3 k+2)+B(3 k-1) \\ & \Rightarrow(3 A+3 B) k+2 A-B=1\end{align} Comparing the cocfficients of $k$ 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 for $A$ and $B$

we get $$A=\dfrac{1}{3}\quad\text{and}\quad B=-\dfrac{1}{3}$$ Thus $$u_n=\dfrac{1}{3}[\dfrac{1}{3 k-1}-\dfrac{1}{3 k+2}]$$ Taking summation of the both sides \begin{align} \sum_{r=1}^n u_n&=\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}{8})\\ &+\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}]\\ S_n&=\dfrac{n}{2(3 n+2)}\end{align}

Find the sum of the series: $\sum_{k=1}^n \dfrac{1}{k^2-k}$

Let $$S_n=\sum_{k=1}^n \dfrac{1}{k^2-k}=\sum_{k=1}^n \dfrac{1}{k(k-1)}$$ Here the $n^{t h}$ term of the series is $$u_n=\dfrac{1}{n(n-1)}$$ Resolving into partial fractions $$\dfrac{1}{k(k-1)}=\dfrac{A}{n}+\dfrac{B}{n-1}$$ Solving the above equation for $A$ and

$B$ we get $A=1$ and $B=-1$. So, \begin{align} u_n&=\dfrac{1}{n}-\dfrac{1}{n-1} \\ S_n&=\sum_{k=1}^n(\dfrac{1}{k}-\dfrac{1}{k-1}) \\ & =(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+\ldots+(\dfrac{1}{n-1}-\dfrac{1}{n})\end{align} Hence the sum is: $$S_n=1-\dfrac{1}{n}=\dfrac{n-1}{n}$$