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 = 23+31221–5 = 611211 = 3
T4 = 7C3 = 35