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:

\[T_n = n\left( n + 1 \right)\left( n + 4 \right)\]

\[ = n\left( n^2 + 5n + 4 \right)\]

\[ = \left( n^3 + 5 n^2 + 4n \right)\]

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

Thus, we have: \[S_n = \sum^n_{k = 1} T_k\]

\[\Rightarrow S_n = \sum^n_{k = 1} k^3 + \sum 5^n_{k = 1} k^2 + \sum 4^n_{k = 1} k \]

\[ \Rightarrow S_n = \sum^n_{k = 1} k^3 + {5\sum}^n_{k = 1} k^2 + {4\sum}^n_{k = 1} k \]

\[ \Rightarrow S_n = \frac{n^2 \left( n + 1 \right)^2}{4} + \frac{5n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + \frac{4n\left( n + 1 \right)}{2}\]

\[ \Rightarrow S_n = \frac{n^2 \left( n + 1 \right)^2}{4} + \frac{5n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + 2n\left( n + 1 \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{n\left( n + 1 \right)}{2} + \frac{5\left( 2n + 1 \right)}{3} + 4 \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{n^2 + n}{2} + \frac{10n + 5}{3} + 4 \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{3 n^2 + 3n + 20n + 10 + 24}{6} \right)\]

\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)}{12}\left( 3 n^2 + 23n + 34 \right)\]