Question

3 + 7 + 14 + 24 + 37 + ...

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: \[S_n = 3 + 7 + 14 + 24 + 37 + . . . + T_{n - 1} + T_n\]  ...(1)

Equation (1) can be rewritten as:

\[S_n = 3 + 7 + 14 + 24 + 37 + . . . + T_{n - 1} + T_n\] ...(2)

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

\[S_n = 3 + 7 + 14 + 24 + 37 + . . . + T_{n - 1} + T_n \]

\[ S_n = 3 + 7 + 14 + 24 + 37 + . . . + T_{n - 1} + T_n \]

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\[0 = 3 + \left[ 4 + 7 + 10 + 13 + . . . + \left( T_n - T_{n - 1} \right) \right] - T_n\]

The sequence of difference of successive terms is 4, 7, 10, 13,...
We observe that it is an AP with common difference 3 and first term 4.
Thus, we have:

\[3 + \left[ \frac{\left( n - 1 \right)}{2}\left\{ 8 + \left( n - 2 \right)3 \right\} \right] - T_n = 0\]

\[ \Rightarrow 3 + \left[ \frac{\left( n - 1 \right)}{2}\left( 3n + 2 \right) \right] - T_n = 0\]

\[ \Rightarrow \left[ \frac{3 n^2 - n + 4}{2} \right] = T_n \]

\[ \Rightarrow \left[ \frac{3}{2} n^2 - \frac{n}{2} + 2 \right] = T_n\]

Now,

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

\[ \therefore S_n = \sum^n_{k = 1} \left( \frac{3}{2} k^2 - \frac{k}{2} + 2 \right) \]

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

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

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

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

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

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