# 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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