Question 715608
SInce the power is {{{17}}}, the expansion would have {{{17+1=18}}} terms.
The degree of all the terms (the sum of the powers of {{{x}}} and of {{{(2y)}}} ) will be {{{17}}}.
The first term will have {{{x^17}}} and {{{(2y)^0}}}.
The second term will have {{{x^16}}} and {{{(2y)^1}}}.
Term number {{{3}}} would have
{{{(2y)^((3-1))=(2y)^2=4y^2}}} as a factor.
As the power is {{{17}}} , it would also have
{{{x^((17-2))=x^15}}} as a factor.
The other factor would be the combinatorial number that tells you how many different groups of 2 you can make with 17 different items:
{{{(matrix(2,1,17,2))=17*16/1/2=136}}}
The third term is
{{{136*x^15*4y^2=highlight(544x^15y^2)}}}