Question

1.2.4 + 2.3.7 +3.4.10 + ...

Answer 1

Let \[T_n\] be the nth term of the given series.

Thus, we have: \[T_n = n\left( n + 1 \right)\left( 3n + 1 \right) = n\left( 3 n^2 + 4n + 1 \right) = \left( 3 n^3 + 4 n^2 + n \right)\]

Now, let 

\[S_n\] be the sum of n terms of the given series.

Thus, we have: 

\[S_n = \sum^n_{k = 1} T_k\]

\[\Rightarrow S_n = \sum 3^n_{k = 1} k^3 + \sum 4^n_{k = 1} k^2 + \sum^n_{k = 1} k \]

\[ \Rightarrow S_n = 3 \sum^n_{k = 1} k^3 + {4\sum}^n_{k = 1} k^2 + \sum^n_{k = 1} k \]

\[ \Rightarrow S_n = \frac{3 n^2 \left( n + 1 \right)^2}{4} + \frac{4n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + \frac{n\left( n + 1 \right)}{2}\]

\[ \Rightarrow S_n = \frac{3 n^2 \left( n + 1 \right)^2}{4} + \frac{2n\left( n + 1 \right)\left( 2n + 1 \right)}{3} + \frac{n\left( n + 1 \right)}{2}\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{3n\left( n + 1 \right)}{2} + \frac{4\left( 2n + 1 \right)}{3} + 1 \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{3 n^2 + 3n}{2} + \frac{8n + 4}{3} + 1 \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{9 n^2 + 9n + 16n + 8 + 6}{6} \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{12}\left( 9 n^2 + 25n + 14 \right)\]