Question

If the constant term in the expansion of
(x3+k/x8)11
is 1320, find k.​

Answer 1

Answer:

The constant term will come from the value of nn, 0≤n≤110≤n≤11, such that

3n=8(11−n).

3n=8(11−n).

That is, n=8n=8. The binomial coefficient for the term

(x3)8⋅(kx8)3=k3

(x3)8⋅(kx8)3=k3

is

11!8!3!=165.

11!8!3!=165.

Given that the constant term equals 13201320, we have 165k3=1320165k3=1320, that is, k3=8k3=8. Hence, k=2k=2.