JEE Main | 2018Differential EquationEasy

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

Create a free account to view solution

View Solution Free
Topic: Differential Equation·Practice all Differential Equation questions

More Differential Equation Questions

The curve in which the slope of the tangent at any point equals the ratio of the abscissa to the ordinate of the point i...The solution of the differential equation y = is -...If A =is a matrix satisfying the equation AAT = 9I, where I is 3 × 3 identity matrix, then the ordered pair (a, b) ...The order of the differential equation whose general solution is given by y = (C1 + C2) cos (x + C3) − C4 , where ...The solution of the equation x + 3y = x is-...