Question 7 & 8 Exercise 5.1

Solutions of Question 7 & 8 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.

Sum to $n$ terms: $1.5 .9+2.6 .10+3.7 .11+\ldots$

The general term of the series is: $T_j=j(j+4)(j+8)$ $$\begin{align} & =j(j^2+12 j+32) \\ & =j^3+12 j^2+32 j\end{align}$$ Taking sum of the both sides of the above equation, we get $$\begin{align} & \sum_{j=1}^n T_j=\sum_{j=1}^n j^3+12 \sum_{j=1}^n j^2+32 \sum_{j=1}^n j \\ & =(\dfrac{n(n+1)}{2})^2+12 \dfrac{n(n+1)(2 n+1)}{6}+32 \dfrac{n(n+1)}{2} \\ & =\dfrac{n^2(n+1)^2}{4}+2 n(n+1)(2 n+1)+16 n(n+1) \\ & =n(n+1)[\dfrac{n(n+1)}{4}+2(2 n+1)+16] \\ & =n(n+1)[\dfrac{n(n+1)+8(2 n+1)+64}{4}] \\ & =\dfrac{n(n+1)}{4} \cdot[n^2+n+16 n+8+64] \\ & =\dfrac{n(n+1)}{4}[n^2+17 n+72] \\ & =\dfrac{n(n+1)}{4}[n^2+9 n+8 n+72] \\ & =\dfrac{n(n+1)}{4}[n(n+9)+8(n+9)] \\ & =\dfrac{n(n+1)(n+8)(n+9)}{4}\end{align}$$

Find the sum of the series upto $2 n$ terms, whose $n^{t h}$-term is $4 n^2+5 n+1$.

Taking summation of the general term of the series $$\begin{align} & \sum_{j=1}^{2 n} T_j=4 \sum_{j=1}^{2 n} j^2+5 \sum_{j=1}^{2 n} j+\sum_{j=1}^{2 n} 1 \\ & =4 \dfrac{2 n(2 n+1)(4 n+1)}{6}+5 \dfrac{2 n(2 n+1)}{2} +2 n \\ & =2 n[4 \dfrac{8 n^2+6 n+1}{6}+\dfrac{5(2 n+1)}{2}+1]\end{align}$$ $$\begin{align} & =2 n[\dfrac{32 n^2+24 n+4+30 n+15+6}{6}] \\ & =\dfrac{n}{3}(32 n^2+54 n+25)\end{align}$$

< Question 6 Question 9 >