The value of the sum $1 \times 2 \times 3+2 \times 3 \times 4+3 \times 4 \times 5+\ldots$ upto $n$ terms is…
The value of the sum $1 \times 2 \times 3+2 \times 3 \times 4+3 \times 4 \times 5+\ldots$ upto $n$ terms is equal to
- $\frac{1}{6} n^2\left(2 n^2+1\right)$
- $\frac{1}{6}\left(n^2-1\right)(2 n-1)(2 n+3)$
- $\frac{1}{8}\left(n^2+1\right)\left(n^2+5\right)$
- $\frac{1}{4} n(n+1)(n+2)(n+3)$
Solution
Let given series be
$S=1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\ldots n$ terms
Now, general term
$
\begin{aligned}
& T_n=\{1+(n-1) \cdot 1\}\{2+(n-1) \cdot 1\}\{3+(n-1) \cdot 1\} \\
& =(1+n-1)(2+n-1)(3+n-1) \\
& =n(n+1)(n+2) \\
& =n\left(n^2+3 n+2\right) \\
& =n^3+3 n^2+2 n
\end{aligned}
$
Now,
$
\begin{aligned}
& \text { Sum of the series } \mathrm{S}=\sum T_n=\sum\left(n^3+3 n^2+2 n\right) \\
& =\sum n^3+3 \sum n^2+2 \sum n \\
& =\frac{n^2(n+1)^2}{4}+\frac{3 n(n+1)(2 n+1)}{6}+\frac{2 n(n+1)}{2} \\
& =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+(2 n+1)+2\right]
\end{aligned}
$
$\begin{aligned} & =\frac{n(n+1)}{2}\left[\frac{n^2+n+4 n+2+4}{2}\right] \\ & =\frac{n(n+1)\left(n^2+5 n+6\right)}{4} \\ & \therefore \mathrm{S}=\frac{n(n+1)(n+2)(n+3)}{4}\end{aligned}$
Asked in: AP EAMCET 2015
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