In the expansion of (1 + 1x) 6, the coefficient of 1x2 is () A. 1B. 6C. 10D. 15

In the expansion of (1 + 1x) 6, the coefficient of 1x2 is () A. 1B. 6C. 10D. 15

The general term of the expansion is tr + 1 = c6rx-r, let - r = - 2 get r = 2, so the coefficient of 1x2 in the expansion is C62 = 15, so the answer is 15

If the coefficient of x ^ 3 in the expansion of (x ^ 2 + 1 / ax) ^ 6 is 5 / 2, then a =?
Please explain in detail what the train of thought is and how the answer is listed

Using binomial theorem
The coefficient of x ^ 3 is C (6,3) / A ^ 3 = 20 / A ^ 3 = 5 / 2
So a = 2

If the coefficient of x2 in the binomial expansion of [x2 + (1 / ax)] 6 is 5 / 2, then a =?

Let K be x ^ 3
Then C6 (k-1) * (x ^ 2) ^ (6-k + 1) * (1 / ax) ^ (k-1)
Index = 2 (6-k + 1) - (k-1) = 3
15-3k=3
k=4
So the coefficient = C63 * (1 / a) ^ 3 = 5 / 2
(1/a)^3=1/8
a=2