Question

How do you find the coefficient of `a^2` in the expansion of `(2a+1)^5`?

Answer 1

Coefficient of `a^2` is 40.

Explanation:

Binomial theorem is

`(x+y)^n = sum_(k=0)^n ((n),(k))x^(n-k)y^k`

We want the a^2 term so if x = 2a then we need `n-k = 2`, since n = 5 this implies that k = 3. Hence the term we are interested in is

`((5),(3))(2a)^(5-3)(1)^3`

`(5!)/(3!2!)*(2a)^2*1^3 = (5*4*(cancel(3*2*1)))/((cancel(3*2*1))(2*1))*4a^2 = 10*4a^2 = 40a^2`

Of course, you could also expand the whole thing out using Pascal's triangle or other methods but this is the easiest way in my opinion.

Answer 2

`40`

Explanation:

`p_n(a)=(2a+1)^n` is a polynomial in `a`

If `b_k` is the coefficient of `a^k` it can be obtained by calculating

`b_k = 1/(k!)(d^k)/(da^k)p_n(a)_{a = 0}`

in the present case

`b_2=1/(2!)(5 xx 4)2^2(2xx0+1)^3=40`