Q. 63.7( 3 Votes )

# Sum the following series to n terms :

1 + 4 + 13 + 40 + 121 + ……….

Answer :

Let s=1 + 4 + 13 + 40 + 121 + …………. + n

By shifting each term by one

S = 1 + 4 + 13 + 40 + 121 + …………. + nth ….(1)

S = 1 + 4 + 13 + 40 + …………. + (n - 1)th + nth …(2)

by (1) - (2) we get

0 = 1 + (3 + 9 + 27 + 81 + …….nth - (n - 1)th - n)

Nth = 1 + (3 + 3^{2} + 3^{3} + 3^{4} + …….nth - (n - 1)th - nth)

Nth = 1 + (3 + 3^{2} + 3^{3} + …….3^{n - 1}) ………(3)

we know

Substituting the above-given value in (3)

thus

s = 1 + 4 + 13 + 40 + 121 + …………. +

We know by property that:

∑ax^{n} + bx^{n - 1} + cx^{n - 2}…….d_{0}=a∑x^{n} + b∑x^{n - 1} + c∑x^{n - 2}…….. + d_{0}∑1

………….. (4)

We know

Thus substituting the above values in(4)

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