====== Question 4 Exercise 5.4 ======
Solutions of Question 4 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 4=====
Find the sum of the series: $\sum_{k=1}^n \dfrac{1}{k^2+7 k+12}$
====Solution====
Let
\begin{align}S_n &=\sum_{k=1}^n \dfrac{1}{k^2+7 k+12} \\
& =\sum_{k=1}^n \dfrac{1}{(k+3)(k+4)}\end{align}
Consider the $n^{\text {th }}$ term of the series
$$u_n=\dfrac{1}{(n+3)(n+4)}$$
Resolving into partial fractions
$$\dfrac{1}{(n+3)(n+4)}=\dfrac{A}{n+3}+\dfrac{B}{n+4}$$
Solving the above equation for $A$ and $B$,

we get $A=1$ and $B=-1$, so
$$u_n=\dfrac{1}{n+3}-\dfrac{1}{n+4}$$
Taking summation of the both sides
\begin{align}S_n&=\sum_{k=1}^n[\dfrac{1}{k+3}-\dfrac{1}{k+4}] \\
& =(\dfrac{1}{4}-\dfrac{1}{5})+(\dfrac{1}{5}-\dfrac{1}{6})+(\dfrac{1}{6}-\dfrac{1}{7}) \\
& +\ldots+(\dfrac{1}{n+3}-\dfrac{1}{n+4})\end{align}
Hence the sum is:
$$S_n=\dfrac{1}{4}-\dfrac{1}{n+4}=\dfrac{n}{4(n+4)}$$


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