Question

1.2.5 + 2.3.6 + 3.4.7 + ...

Answer 1

Let $T_n$ be the nth term of the given series.

Thus, we have:

$$\begin{gathered} T_n=n\left(n+1\right)\left(n+4\right) \\ =n\left(n^2+5n+4\right) \\ =\left(n^3+5n^2+4n\right) \end{gathered}$$

Now, let $S_n$ be the sum of n terms of the given series.

Thus, we have:
$S_n=\sum _{k=1}^n{T}_k$

$$\begin{gathered} \Rightarrow S_n=\sum _{k=1}^n{k}^3+\underset{k=1}{\overset{n}{\sum 5}}{k}^2+\underset{k=1}{\overset{n}{\sum 4}}k \\ \Rightarrow S_n=\sum _{k=1}^n{k}^3+\underset{k=1}{\overset{n}{5\sum }}{k}^2+\underset{k=1}{\overset{n}{4\sum }}k \\ \Rightarrow S_n=\frac{n^2{\left(n+1\right)}^2}{4}+\frac{5n\left(n+1\right)\left(2n+1\right)}{6}+\frac{4n\left(n+1\right)}{2} \\ \Rightarrow S_n=\frac{n^2{\left(n+1\right)}^2}{4}+\frac{5n\left(n+1\right)\left(2n+1\right)}{6}+2n\left(n+1\right) \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left(\frac{n\left(n+1\right)}{2}+\frac{5\left(2n+1\right)}{3}+4\right) \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left(\frac{n^2+n}{2}+\frac{10n+5}{3}+4\right) \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2}\left(\frac{3n^2+3n+20n+10+24}{6}\right) \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{12}\left(3n^2+23n+34\right) \end{gathered}$$