# Question 6 Exercise 5.1

Solutions of Question 6 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 6

Sum: $1.2 \cdot 3+2 \cdot 3.4+3.4 .5+\ldots$ to $n$-terms.

### Solution

We see that each term of the given series is the product of corresponding terms of the three series $1+2+3+\ldots, \quad 2+3+4+5+\ldots$ and $3+4+5+6+7+\ldots$

whose $n^{t h}$ terms are $j, j+1$ and $j+2$ respectively, therefore the $n^{t h}$ term of the given series is: \begin{align} & T_j=j(j+1)(j+2)-j(j^2+3 j+2) \\ & =j^3+3 j^2+2 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+3 \sum_{j=1}^n j^2+2 \sum_{j=1}^n j \\ & =(\dfrac{n(n+1)}{2})^2+3 \dfrac{n(n+1)(2 n+1)}{6}+2 \dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)}{2}[\dfrac{n(n+1)}{2}+3 \dfrac{(2 n+1)}{3}+2] \\ & =\dfrac{n(n+1)}{2}[\dfrac{n^2+n}{2}+\dfrac{6 n+3}{3}+2] \\ & =\dfrac{n(n+1)}{2}[\dfrac{3 n^2+3 n+12 n+6+12}{6}] \\ & =\dfrac{n(n+1)}{12}[3 n^2+15 n+18] \\ & =\dfrac{3 n(n+1)}{12}[n^2+5 n+6] \\ & =\dfrac{n(n+1)(n^2+5 n+6)}{4} \\ &=\dfrac{n(n+1)(n^2+3n+2 n+6)}{4}\\ &=\dfrac{n(n+1)(n(n+3)+2( n+3))}{4}\\ &=\dfrac{n(n+1)(n+2) (n+3)}{4}\end{align}

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