# Question 1 Exercise 5.3

Solutions of Question 1 of Exercise 5.3 of Unit 05: Mascellaneous 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 $n$ term and sum to $n$ terms each of the series $4+13+28+49+76+\ldots$

\begin{align} & a_2-a_1=13-4=9 \\ & a_3-a_2=28-13=15 \\ & a_4-a_3=49-28=21 \\ & \cdots \quad \cdots \quad \cdots \\ & \cdots \quad \cdots \quad \cdots \\ & a_n-a_{n-1}=(n-1)th \quad\text{term of sequence}\quad 9,15,21,...\end{align} Which is in A.P. column wise, we get \begin{align} a_n-a_{n-1}&=9+15+21+\ldots+(n-1)\\ & =\dfrac{n-1}{2}[2.9+(n-2) \cdot 6] \\ & =\dfrac{n-1}{2}[18+6 n-12] \\ & =\dfrac{n-1}{2}[6+6 n] \\ \Rightarrow a_n&=3(n^2-1)+a_1 \\ \Rightarrow a_n&=3 n^2-3+4 \because a_1=4 \\ \Rightarrow a_n&=3 n^2+1 \end{align} Taking summation of the both sides \begin{align} & \sum_{r=1}^n a_r=3 \sum_{r=1}^n r^2+\sum_{r=1}^n 1 \\ & =3 \dfrac{n(n+1)(2 n+1)}{6}+n \\ & =n \cdot[\dfrac{(n+1)(2 n+1)}{2}+1] \\ & =n \cdot[\dfrac{2 n^2+3 n+1+2}{2}] \\ & \Rightarrow \sum_{r=1}^n a_r=\dfrac{n}{2}(2 n^2+3 n+3)\\ \text{Hence} \quad 3 n^2+1&;\dfrac{n}{2}(2 n^2+3 n+3)\end{align}