Question

Find the sum of the series whose nth term is:

n³ − 3ⁿ

Answer 1

Let \[T_n\] be the nth term of the given series.
Thus, we have:

\[T_n = n^3 - 3^n\]

Let \[S_n\] be the sum of n terms of the given series.
Now, \[S_n = \sum^n_{k = 1} T_k\]

\[\Rightarrow S_n = \sum^n_{k = 1} \left( k^3 - 3^k \right)\]

\[ \Rightarrow S_n = \sum^n_{k = 1} k^3 - \sum^n_{k = 1} 3^k \]

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

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

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