Question

2 + 4 + 7 + 11 + 16 + ...

Answer 1

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

Thus, we have:

\[S_n = 2 + 4 + 7 + 11 + 16 + . . . + T_{n - 1} + T_n\]   ...(1)

Equation (1) can be rewritten as:

\[S_n = 2 + 4 + 7 + 11 + 16 + . . . + T_{n - 1} + T_n\]   ...(2)

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

\[S_n = 2 + 4 + 7 + 11 + 16 + . . . + T_{n - 1} + T_n \]

\[ S_n = 2 + 4 + 7 + 11 + 16 + . . . + T_{n - 1} + T_n \]

\[ -        \   \  - - - - -      \  -  \ -  \]

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\[ 0 = 2 + \left[ 2 + 3 + 4 + 5 + 6 + . . . + \left( T_n - T_{n - 1} \right) \right] - T_n\]

\[\Rightarrow 2 + \left[ \frac{\left( n - 1 \right)}{2}\left( 4 + \left( n - 2 \right)1 \right) \right] - T_n = 0\]

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

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

\[ \Rightarrow \left[ \frac{n^2}{2} + \frac{n}{2} + 1 \right] = T_n\]

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

\[ \therefore S_n = \sum^n_{k = 1} \left( \frac{k^2}{2} + \frac{k}{2} + 1 \right)\]

\[ = \frac{1}{2} \sum^n_{k = 1} k^2 + \frac{1}{2} \sum^n_{k = 1} k + \sum^n_{k = 1} 1\]

\[ = \frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{12} + \frac{n\left( n + 1 \right)}{4} + n\]

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

\[ = \frac{n}{12}\left( 2 n^2 + 6n + 16 \right)\]

\[ = \frac{n}{6}\left( n^2 + 3n + 8 \right)\]