If the coefficient sum of (x + A / x) (2x-1 / x) expansion term is 2, then the constant term is?

If the coefficient sum of (x + A / x) (2x-1 / x) expansion term is 2, then the constant term is?

(X+a/X)(2X-1/X)=2x^2-1+2a-a/x^2
Let x = 1, the sum of coefficients of the expansion is 2-1 + 2a-a = 1 + a = 2
The solution is a = 1, and the expansion is 2x ^ 2 + 1-1 / x ^ 2,
So the constant term is 1

The coefficients and constants of x ^ 4 in the expansion of (1-x ^ 2) (2x + 1) ^ 5

In the (2x + 1) ^ 5 expansion, the multiplication of the term of x ^ 2 by (- x ^ 2) and the multiplication of x ^ 4 by 1 determine the coefficient of x ^ 4 in the expansion
Then in the (2x + 1) ^ 5 expansion
The item of x ^ 2 is C (5,3) * (2x) ^ 2 * 1 ^ 3 = 10 * 4x ^ 2 * 1 = 40x ^ 2
The item of x ^ 4 is C (5,1) * (2x) ^ 4 * 1 ^ 1 = 5 * 16x ^ 4 * 1 = 80x ^ 4
Therefore, in the expansion, the coefficient of x ^ 4 = 80 * 1 + 40 * (- 1) = 40
The constant term is the same as the method
Constant term = C (5,5) * (2x) ^ 0 * 1 ^ 5 = 1 * 1 * 1 = 1

Coefficient of x ^ 6 term in the expansion of (x ^ 1 / 2 + 1) ^ 6 (2x + 1) ^ 5

The coefficients of the term x &# 8310; in the expansion of [(√ x) + 1] &# 8310; (2x + 1) &# 8309
The original formula = [x & # 179; + 6x ^ (5 / 2) + 15x & # 178; + 20x ^ (3 / 2) + 15x + 6x ^ (1 / 2) + 1] [32x & # 8309; + 80x & # 8308; + 80x & # 179; + 40x & # 178; + 10x + 1]
Therefore, the term containing X & # - 8310; consists of the following terms: (1 × 80 + 15 × 80) x & # - 8310; = 1280 X & # - 8310;
That is to say, the expansion contains X & # 8310; and the coefficient is 1280