Question 1187158

For any value of {{{n}}}, whether positive, negative, integer or non-integer, the value of the {{{n}}}th power of a binomial is given by:

 {{{(a+b)^n}}}={{{sum(C(n,k)a^(n-k)b^k,0,n)}}}, where {{{C(n,k)=n!/((n-k)!k!)}}} and {{{n!}}}={{{1*2}}}*…*{{{n}}}.


Now, calculate the product for every value of {{{k}}} from {{{0}}} to {{{8}}}.

{{{k=0}}}: {{{C(8,0)(2x^2)^(8-0)*1^0=(8!/((8-0)!0!))*(2x^2)^(8-0)*1^0=256x^16}}}

{{{k=1}}}: {{{C(8,1)(2x^2)^(8-1)*1^1=(8!/((8-1)!1!))*(2x^2)^(8-1)*1^1=1024x^14}}}

{{{k=2}}}: {{{C(8,2)(2x^2)^(8-2)*1^2=(8!/(8-2)!2)!(2x^2)^(8-2)*1^2=1792x^12}}}

so, your answer is{{{ 1792x^12}}}