The binomial expansion of (3+2x)4(3 + 2x)^4(3+2x)4 is given by
(3+2x)4=A+Bx+216x2+Dx3+16x4(3 + 2x)^4 = A + Bx + 216x^2 + Dx^3 + 16x^4(3+2x)4=A+Bx+216x2+Dx3+16x4
Find the value of AAA and the value of BBB.
Show that
(3+2x)4−(3−2x)4=Cx+Dx3(3 + 2x)^4 - (3 - 2x)^4 = Cx + Dx^3(3+2x)4−(3−2x)4=Cx+Dx3
where CCC and DDD are constants to be found.
Hence, or otherwise, find
∫((3+2x)4−(3−2x)4)dx\int \left( (3 + 2x)^4 - (3 - 2x)^4 \right) dx∫((3+2x)4−(3−2x)4)dx