Solveeit Logo

Question

Question: If sum of all the coefficients in the expansion of (x<sup>3/2</sup> + x<sup>–1/3</sup>)<sup>n</sup> ...

If sum of all the coefficients in the expansion of (x3/2 + x–1/3)n is 128, then the coefficient of x5 is –

A

35

B

45

C

7

D

None of these

Answer

35

Explanation

Solution

Sum of all coefficients,

Put x = 1, 2n = 128 0 Ž n = 7

Now in (x3/2 + x–1/3)7, r = 212532+13\frac{\frac{21}{2}–5}{\frac{3}{2} + \frac{1}{3}} = 112116\frac{\frac{11}{2}}{\frac{11}{6}} = 3

T4 = 7C3 = 35