JEE Main | 2018Differential EquationHard

Question

The sum of the co-efficients of all odd degree terms in the expansion of x+x3-15+x-x3-15, (x > 1) is-

Options

A.

0

B.

1

C.

2

D.

-1

Solution

using (x + a)5 + (x – a)5
= 2[5C0 x5 + 5C2 x3·a2 + 5C4x·a4]
x+x3-15+x-x3-15
= 2[5C0 x5 + 5C2 x3(x3 - 1) + 5C4 x(x3 - 1)2]
⇒ 2[x5 + 10x6 - 10x3 + 5x7 - 10x4 + 5x]
considering odd degree terms,
2[x5 + 5x7 - 10x3 + 5x]
∴Sum of coefficients of odd terms is 2

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