====== 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}




====Go To====
<text align="left"><btn type="primary">[[math-11-kpk:sol:unit05:ex5-1-p3 |< Question 4 & 5 ]]</btn></text>
<text align="right"><btn type="success">[[math-11-kpk:sol:unit05:ex5-1-p5|Question 7 & 8 >]]</btn></text>