# Question 4 & 5 Exercise 5.1

Solutions of Question 4 & 5 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.

Find the sum of the $2+(2+5)+(2+5+8)+\ldots$ up to $n$ terms.

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

Sum: $2+5+10+17+\ldots$ to $n$ terms.

First we reform the given series as: $$(1+1^2)+(1+2^2)+(1+3^2)+(1+4^2)+\ldots$$ Hence the general term of the series is: $T_j=1+j^2$

Taking sum of the both sides of the above equation, we get \begin{align} & \sum_{j=1}^{j=n} T_j=\sum_{j=1}^{j=n} 1+\sum_{j=1}^{j=n} j^2 \\ & =n+\dfrac{n(n+1)(2 n+1)}{6} \\ & =n[\dfrac{6+(n+1)(2 n+1)}{6}] \\ & =n[\dfrac{6+2 n^2+3 n+1}{6}] \\ & =\dfrac{n}{6}(2 n^2+3 n+7)\end{align}