Question 2 Exercise 5.3

Solutions of Question 2 of Exercise 5.3 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 $n$ term and sum to $n$ terms each of the series $4+14+30+52+80+114+\ldots$

\begin{align} & a_2-a_1=14-4=10 \\ & a_3-a_2=30-14=16 \\ & a_4-a_3=52-30=22 \\ & \cdots \quad \cdots \quad \cdots \\ & \cdots \quad \cdots \quad \cdots \\ & a_n-a_{n-1}=(\mathrm{n}-1)\text{ term of the sequence} 10,16,22, \ldots\end{align} which is a A.P. Adding column wise, we get \begin{align} a_n-a_1&=10+16+22+\ldots+(n-1) \text { terms } \\ & =\dfrac{n-1}{2}[2 \cdot 10+(n-2) \cdot 6] \\ & =\dfrac{n-1}{2}[20+6 n-12] \\ & =\dfrac{n-1}{2}[6 n+8] \\ & =2 \cdot \dfrac{n-1}{2}[3 n+4] \\ \Rightarrow a_n-a_1&=(n-1)(3 n+4) \\ \Rightarrow a_n&=3 n^2+n-4+a_1 \\ \Rightarrow a_n&=3 n^2+n-4+4 \quad \because a_1=4 \\ \Rightarrow a_n&=3 n^2+n\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 r \\ & =3 \cdot \dfrac{n(n+1)(2 n+1)}{6}+\dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)}{2}[3 \dfrac{2 n+1}{3}+1] \\ & =\dfrac{n(n+1)}{2}[2 n+2] \\ & =\dfrac{2 n(n+1)^2}{2} \\ & \Rightarrow \sum_{r=1}^n a_r=n(n+1)^2\\ \text{Hence}\quad 3 n^2+n&;n(n+1)^2\end{align}

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