====== 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.

=====Question 2=====
Find sum of the series: $\sum_{k=1}^n \dfrac{1}{9 k^2+3 k-2}$
====Solution====
\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}

=====Question 3=====
Find the sum of the series: $\sum_{k=1}^n \dfrac{1}{k^2-k}$
====Solution====
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:
$$S_n=1-\dfrac{1}{n}=\dfrac{n-1}{n}$$




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