Question

1 + 3 + 6 + 10 + 15 + ...

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: 

\[S_n = 1 + 3 + 6 + 10 + 15 + . . . + T_{n - 1} + T_n\]  ....(1)

Equation (1) can be rewritten as:

\[S_n = 1 + 3 + 6 + 10 + 15 + . . . + T_{n - 1} + T_n\]  ...(2)

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

\[S_n = 1 + 3 + 6 + 10 + 15 + . . . + T_{n - 1} + T_n \]

\[ S_n = 1 + 3 + 6 + 10 + 15 + . . . + T_{n - 1} + T_n \]

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

The sequence of difference of successive terms is 2, 3, 4, 5,...
We observe that it is an AP with common difference 1 and first term 2.
Thus, we have:

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

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

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

Now,

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

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

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

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

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

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

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

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