# Question 9 Exercise 5.1

Solutions of Question 9 of Exercise 5.1 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 9(i)

Find the sum of the $n$ terms of the series whose $n$-term is (32 n^2+54 n+25).

### Solution

The $n$-term of the the series is given as: \begin{align} & T_n=n^2(2 n+3)=2 n^3+3 n^2 \\ & \Rightarrow T_j=2 j^3+3 j^2\end{align} Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=2 \sum_{j=1}^n j^3+3 \sum_{j=1}^n j^2 \\ & =2(\dfrac{n(n+1)}{2})^2+3 \dfrac{n(n+1)(2 n+1)}{6} \\ & =\dfrac{n(n+1)}{2}[2 \dfrac{n(n+1)}{2}+3 \dfrac{2 n+1}{3}] \\ & =\dfrac{n(n+1)}{2}[n^2+n+2 n+1] \\ & =\dfrac{n(n+1)}{2}(n^2+3 n+1)\end{align}

## Question 9(ii)

Find the sum of the $n$ terms of the series whose $n$-term is $3\left(4^n+2 n\right)-4 n^3$.

### Solution

The general term of the series is: $$T_j=3(4^j+2 j^2)-4 n^3$$ Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^n T_j=3 \sum_{j=1}^n 4^j+6 \sum_{j=1}^n j^2-4 \sum_{j=1}^n j^3 \\ & =3 \dfrac{4(4^n-1)}{4-1}+6 \dfrac{n(n+1)(2 n+1)}{6}-4 \dfrac{n^2(n+1)^2}{4} \\ & =4(4^n-1)+n(n+1)[2 n+1-n(n+1)]\\ &=4^{n+1}-4+n(n+1)(-n^2+ n+1)\\ &=4^{n+1}-4-n(n+1)(n^2- n-1)\end{align}

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