Find the coefficient of w^5 in the expansion of (1+2w)^10?

Answer 1

C(5,10) = 10!/5!5! = 10*9*8*7*6/120=

=9*8*7*6/12 = 9*4*7 = 252.

This must be multiplied by 2^5, because of 2w, so

252*32 = 8064

Answer 2

The coefficient of x^t in the expression (1+x)^n is n!/t!(n-t)!
so in your case the answer is 10!/5!5! (2w)^5
=252 * 2^5 * w^5 = 8064

Answer 3

n=10
using the binomial expansion;
let x=2w
(1+x)^10
=1+10x+90x^2/2+720x^3/6
+5040x^4/24+30240x^5/120+....
=1+10x+45x^2+120x^3
+210x^4+252x^5+.......+
the coefficient of x^5=252
however,x=2w
(2w)^5=2^5*w^5
hence,coefficient of
w^5 in the expansion
(1+2w)^10
=2^5*252=8064

i hope that this helps

Answer 4

It is 8064