Find the coefficient of x^10 in the expansion (2x + 1÷x)^100
Let U = (2x + 1/x)^100
Then, U = x^(-100)(2x^2 + 1)^100
U = x^(-100)(1 + v)^n, where n = 100 and v = 2x^2
Using the Binomial theorem to expand the exponent term,
U = x^(-100){1 + n.v + n(n-1)/2.v^2 + n(n-1)(n-2)/3!.v^3 + … n!/[r!(n – r)!].v^r + ...}
Where
t_r = nCr.v^r, r = 0,1,2,... , nCr = n!/[r!(n – r)!], v = 2x^2
t_r = nCr.2^r.x^(2r)
For coefficient of x^10, we need 2r = 10 + 100 = 110 ==> r = 55
Then t_55 = x^(-100)*100C55.2^55.x^(110)
t_55 = 100C55.2^55.x^(10)
t_55 = 2^55.100C55.x^10
Coefficient of x^10 is: 2^55.100C55
t_55 is the 56th term in the expansion