What is the coefficient of x3y2 in the expansion of (2x + y)5?

Step-by-step

The coefficients of terms of (p+q)^n can be found by the Pascal's triangle for small values of n.  Pascal's triangle will start with (1,1)  = coefficients of (p,q)^n =1.  For n=2, we add successive terms of the previous value of n.  Thus for n-2, we have (, 1+1,11=(1,2,1), for n=3, we have (1,3,3,1), giving the following pattern:

(1,1)

(1,2,1)

(1,3,3,1)

(1,4,6,4,1)

(1,5,10,10,5,1)

meaning for n=5, the binomial expansion for (P+Q)^5 is

P^5+5P^4Q+10P^3Q^2+10P^2Q^3+5PQ^4+Q^5

Setting P=2x, Q=y in the term 10P^3Q^2, we get a term

10(2x)^3(y)^2

=10(8x^3)(y^2)

=80x^3y^2

So the required coefficient is K=80.

We can also find the coefficient 10 by binomial expansion of

n=5, x=3 in

C(n,x) = n! / (x! (n-x)!) = 5! / (2!3!) = 5*4*3/(1*2*3) = 10

Then again substituting 10(2x)^3(y)^2 = 80x^3y^2

to get the coefficient K=80.

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