Question

1 + 3 + 7 + 13 + 21 + ...

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:

\[S_n = 1 + 3 + 7 + 13 + 21 + . . . + T_{n - 1} + T_n\]    ...(1)

Equation (1) can be rewritten as:

\[S_n = 1 + 3 + 7 + 13 + 21 + . . . + T_{n - 1} + T_n\]   ...(2)
On subtracting (2) from (1), we get:

\[S_n = 1 + 3 + 7 + 13 + 21 + . . . + T_{n - 1} + T_n \]

\[ S_n = 1 + 3 + 7 + 13 + 21 + . . . + T_{n - 1} + T_n \]

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\[0 = 1 + \left[ 2 + 4 + 6 + 8 + . . + \left( T_n - T_{n - 1} \right) \right] - T_n = 0\]

The sequence of difference of successive terms is 2, 4, 6, 8,...
We observe that it is an AP with common difference 2 and first term 2.
Thus, we have:

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

\[ \Rightarrow 1 + \left[ n^2 - n \right] = T_n \]

\[ \Rightarrow \left[ n^2 - n + 1 \right] = T_n\]

Now,

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

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

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

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

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

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

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