3 + 5 + 9 + 15 + 23 + ………….

Solution:

Let T_n and S_n be nth term and the sum of n terms respectively, of the given series.

Then, according to the question:

$ {{S}_{n}}~=\text{ }3\text{ }+\text{ }5\text{ }+\text{ }9\text{ }+\text{ }15\text{ }+\text{ }23\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{ }3\text{ }+\text{ }5\text{ }+\text{ }9\text{ }+\text{ }15\text{ }+\text{ }23\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}}~\ldots \ldots ..\left( 2 \right) $

Subtracting equation (2) from (1) we have:

$ {{S}_{n}}~=\text{ }3\text{ }+\text{ }5\text{ }+\text{ }9\text{ }+\text{ }15\text{ }+\text{ }23\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}} $

$ {{S}_{n}}~=\text{ }3\text{ }+\text{ }5\text{ }+\text{ }9\text{ }+\text{ }15\text{ }+\text{ }23\text{ }+\text{ }\ldots \ldots \ldots \ldots .\text{ }+\text{ }{{T}_{n-1}}~+\text{ }{{T}_{n}} $

$ 0\text{ }=\text{ }3\text{ }+\text{ }\left[ 2\text{ }+\text{ }4\text{ }+\text{ }6\text{ }+\text{ }8\text{ }+\text{ }\ldots \text{ }+\text{ }\left( {{T}_{n}}-{{T}_{n-1}} \right) \right]-{{T}_{n}} $

The differences between the successive terms are as follows:

5-3 = 2, 9-5 = 4, 15-9 = 6,

As we can see that these differences are in A.P.

We have:

Therefore, the sum of the series is n/3 (n² + 8)