Question

2 + 5 + 10 + 17 + 26 + ...

 

Answer 1

Let  \[T_n\]  be the nth term and \[S_n\]  be the sum of n terms of the given series.

Thus, we have:

\[S_n = 2 + 5 + 10 + 17 + 26 + . . . + T_{n - 1} + T_n\]    ...(1) 

Equation (1) can be rewritten as: 

\[S_n = 2 + 5 + 10 + 17 + 26 + . . . + T_{n - 1} + T_n\]  ...(2) 

 

On subtracting (2) from (1), we get:

\[S_n = 2 + 5 + 10 + 17 + 26 + . . . + T_{n - 1} + T_n \]
\[ S_n = 2 + 5 + 10 + 17 + 26 + . . . + T_{n - 1} + T_n \]
\[ 0 = 2 + \left[ 3 + 5 + 7 + 9 + . . . + \left( T_n - T_{n - 1} \right) \right] - T_n\]

The sequence of difference of successive terms is 3, 5, 7, 9,...
We observe that it is an AP with common difference 2 and first term 3.
Thus, we have:

\[2 + \left[ \frac{\left( n - 1 \right)}{2}\left\{ 6 + \left( n - 2 \right)2 \right\} \right] - T_n = 0\]
\[ \Rightarrow 2 + \left[ n^2 - 1 \right] = T_n \]
\[ \Rightarrow \left[ n^2 + 1 \right] = T_n\]

Now,

\[\because S_n = \sum^n_{k = 1} T_k \]
\[ \therefore S_n = \sum^n_{k = 1} \left( k^2 + 1 \right) \]
\[ \Rightarrow S_n = \sum^n_{k = 1} k^2 + \sum^n_{k = 1} 1\]
\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + n\]
\[ \Rightarrow S_n = \frac{n\left( n + 1 \right)\left( 2n + 1 \right) + 6n}{6}\]
\[ \Rightarrow S_n = \frac{n\left( 2 n^2 + 3n + 7 \right)}{6}\]