# Question 2 & 3 Review Exercise

Solutions of Question 2 & 3 of Review Exercise 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.

Sum the series to $n$ terms $1.2+2.3+3.4+\ldots$

The $n^{\text {th }}$ term is: $$a_n=n(n+1)=n^2+n$$ Taking summation of the both sides \begin{align} \sum_{r=1}^n a_r&=\sum_{r=1}^n r^2+\sum_{r=1}^n r \\ & =\dfrac{n(n+1)(2 n+1)}{6}+\dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)}{2}[\dfrac{2 n+1}{3}+1] \\ & =\dfrac{n(n+1)}{2} \cdot \dfrac{2 n+1+3}{3} \\ & =\dfrac{n(n+1)}{2} \cdot \dfrac{2 n+4}{3} \\ & =\dfrac{n(n+1)(n+2)}{3} \\ \therefore S_n &=\sum_{r=1}^n a_r=\dfrac{n(n+1)(n+2)}{3}\end{align}

Sum the series: $1.3 .5+2.4 .6+3.5 .7+\ldots$ to $n$ terms.

In the given series each term is the product of corresponding terms of the sequences $1,2.3, \ldots, 3,4,5 \ldots$ and $5,6,7, \ldots$

Each one have $n^{t h}$ term $n, \quad n+2$ and $n+4$ respectively.

Therefore, $n^{t h}$ term of the series is $a_n=n(n+2)(n+4)$ or $a_n=n^3+6 n^2+8 n$

Taking summation of the both sides \begin{align} \sum_{r=1}^n a_r&=\sum_{r-1}^n r^3+6 \sum_{r-1}^n r^2+8 \sum_{r=1}^n r \\ & =[\dfrac{n(n+1)}{2}]^2+6 \dfrac{n(n+1)(2 n+1)}{6}+8 \dfrac{n(n+1)}{2} \\ & =\dfrac{n^2(n+1)^2}{4}+n(n+1)(2 n+1)+ 4 n(n+1) \\ & =n(n+1) [\dfrac{n(n+1)}{4}+(2 n+1)+4] \\ & =n(n+1)[\dfrac{n^2+n+8 n+4+16}{4}] \\ & =\dfrac{n(n+1)(n^2+9 n+20)}{4}\\ & =\dfrac{n(n+1)(n^2+4n+5 n+20)}{4}\\ & =\dfrac{n(n+1)(n(n+4)+5(n+4)}{4}\\ & =\dfrac{n(n+1)(n+4)(n+5)}{4}\end{align} Hence sum of the $n$ terms of the series is: $S_n=\dfrac{n(n+1)(n+4)(n+5)}{4}$.