====== 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.

=====Question 2===== 
Find $n$ term and sum to $n$ terms each of the series $4+14+30+52+80+114+\ldots$
====Solution====
\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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