====== 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.

=====Question 4=====
Find the sum of the $2+(2+5)+(2+5+8)+\ldots$ up to $n$ terms.
====Solution====
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}

=====Question 5=====
Sum: $2+5+10+17+\ldots$ to $n$ terms.
====Solution====
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}




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