Expand (2+3x)^5 in ascending power of x, up to x^3.

We'll apply Binomial
Theorem:

(a+b)^n = C(n,0)*a^n + C(n,1)*a^(n-1)*b + ... +
C(n,k)*a^(n-k)*b^k + .... + C(n,n)*b^n

We'll put a = 2, b =
3x and n = 5

(2+3x)^5 = C(5,0)*2^5 + C(5,1)*2^4*3x +
C(5,2)*2^3*9x^2 + C(5,3)*2^2*27x^3 + ...

C(5,0) =
1

C(5,1) = 5

C(5,2) = 5(5-1)/2
= 10 = C(5,3)

(2+3x)^5 = 32 + 5*16*3x + 10*8*9x^2 +
10*4*27x^3 + ...

(2+3x)^5 = 32 + 240x +
720x^2 + 1080x^3 + ...