Question 1203588

To find the coefficient of the {{{5}}}th term in the expansion of {{{(2x^2+ 3y^3)^10}}}, we need to use the binomial theorem. 

The general term in the expansion of {{{(2x^2+ 3y^3)^10}}} is given by:


{{{T(r+1) = C(10, r) * (2x^2)^(10-r) * (3y^3)^r}}}


where {{{C(10, r)}}} is the binomial coefficient of {{{10}}} and {{{r}}}, given by:


{{{C(10, r) = 10! / (r! * (10 - r)!)}}}


To find the {{{5}}}th term, we need to substitute {{{r = 4}}}, since the first term has {{{r = 0}}}. Therefore, the {{{5}}}th term is:


{{{T(5) = C(10, 4) * (2x^2)^(10-4) * (3y^3)^4}}}

= {{{C(10, 4) * (2x^2)^6 * (3y^3)^4}}}

= {{{210 * 64x^12 * 81y^12}}}

= {{{1088640x^12*y^12}}}


Therefore, the coefficient of the {{{5}}}th term in the expansion of {{{(2x^2 + 3y^3)^10}}} is {{{1088640}}}.