1 + 3 + 7 + 13 + 21 + …
Solution:
Let Tn represent the nth term and let Sn represent the sum to n terms of the given series.
According to the question:
$ {{S}_{n}}~=\text{ }1\text{ }+\text{ }3\text{ }+\text{ }7\text{ }+\text{ }13\text{ }+\text{ }21\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}}~\ldots \text{ }\left( 1 \right) $
We can rewrite equation (1) as:
$ {{S}_{n}}~=\text{ }1\text{ }+\text{ }3\text{ }+\text{ }7\text{ }+\text{ }13\text{ }+\text{ }21\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}}~\ldots \ldots ..\left( 2 \right) $
Upon subtracting equation (2) from (1) we get
$ {{S}_{n}}~=\text{ }1\text{ }+\text{ }3\text{ }+\text{ }7\text{ }+\text{ }13\text{ }+\text{ }21\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}} $
$ {{S}_{n}}~=\text{ }1\text{ }+\text{ }3\text{ }+\text{ }7\text{ }+\text{ }13\text{ }+\text{ }21\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}} $
$ 0\text{ }=\text{ }1\text{ }+\text{ }\left[ 2\text{ }+\text{ }4\text{ }+\text{ }6\text{ }+\text{ }8\text{ }+\text{ }\ldots \text{ }+\text{ }\left( {{T}_{n}}~\text{ }{{T}_{n-1}} \right) \right]-{{T}_{n}} $
The successive terms differ by 2, 4, 6, 8. As we can see that these differences are in A.P
Now, we have:
Therefore, the sum of the series is n/3 (n2 + 2)