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Sum the following series to n terms:
CBSE | Class 11 | Exercise 21.2 | Maths | RD Sharma | Some Special Series
Sum the following series to n terms:

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:

RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series image - 10

Therefore, the sum of the series is n/3 (n2 + 2)

i

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