Question

The sum of the nth term of the series 1.2.5 + 2.3.6 + 3.4.7 + .....n terms is $\frac{1}{12}n(n+1)(3n^2+23n+34)$

• A. True
• B. False

Answer 1

The correct option is A True

We have,

$\mathrm{1.2.5}+\mathrm{2.3.6}+\mathrm{3.4.7}+.......+nterms$

$1(1+1)(1+4)+2(2+1)(2+4)+3(3+1)(3+4)+............+nterms$

Therefore, the general term

$\Rightarrow n(n+1)(n+4)$

Now, the sum of this series will be

$\Rightarrow \sum _{k=1}^{n}k(k+1)(k+4)$

$\Rightarrow \sum _{k=1}^n(k^3+5k^2+4k)$

Therefore,

$\Rightarrow \frac{\left(n\right(n+1){)}^2}{4}+\frac{5n(n+1)(2n+1)}{6}+\frac{4n(n+1)}{2}$

$\Rightarrow \frac{n(n+1)}{2}(\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4)$

$\Rightarrow \frac{n(n+1)}{2}\left(\frac{3n^2+23n+34}{6}\right)$

$\Rightarrow \frac{n(n+1)}{12}\left(3n^2+23n+34\right)$

Hence, this is the answer.