Browse Questions

# Find the sum of $n$ terms of the series whose $n^{th}$ term is given by $n(n+1)(n+4)$

$\begin{array}{1 1}\large\frac{n(n+1)(3n^2+23n+34)}{24} \\ \large\frac{n(n+1)(3n^2+23n+24)}{12} \\ \large\frac{n(n+1)(3n^2+20n+34)}{12}\\\large\frac{n(n+1)(3n^2+23n+34)}{12}\end{array}$

Toolbox:
• Sum of $n$ terms of any series =$S_n=\sum t_n$
• $\sum (A+B+C)=\sum A+\sum B+\sum C$
• $\sum kA=k.\sum A$ where $k$ is a constant.
• $\sum n=\large\frac{n(n+1)}{2}$
• $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$
• $\sum n^3=\large\frac{n^2(n+1)^2}{4}$
Given: $n^{th}$ term of the series is $t_n=n(n+1)(n+4)$
We know that sum of $n$ terms of any series =$S_n=\sum t_n$
$\Rightarrow\:S_n=\sum t_n=\sum n(n+1)(n+4)$
$\qquad\:=\sum( n^3+5n^2+4n)$
We know that $\sum (A+B+C)=\sum A+\sum B+\sum C$
$\Rightarrow\:S_n=\sum n^3+\sum 5n^2+\sum 4n$
We know that $\sum kA=k.\sum A$ where $k$ is a constant.
$\qquad=\sum n^3+5\sum n^2+4\sum n$
$\qquad =\large\frac{n^2(n+1)^2}{4}$$+5\times \large\frac{n(n+1)(2n+1)}{6}$$+4\times \large\frac{n(n+1)}{2}$
$\qquad=\large\frac{n(n+1)}{2}$$\bigg[\large\frac{n(n+1)}{2}$$+\large\frac{5(2n+1)}{3}$$+4\bigg] \qquad=\large\frac{n(n+1)}{2}$$\bigg[\large\frac{3n^2+3n+20n+10+24}{6}\bigg]$
$\qquad=\large\frac{n(n+1)(3n^2+23n+34)}{12}$