How do you find the coefficient of #x^6# in the expansion of #(x^2+4)^10#?

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Answer 1 · 2016-08-04T11:53:40

#1966080#

#(x^2+4)^10 equiv (y + 4)^10# after the substitution #y = x^2#

Now, looking at the expansion we will have

#(y + 4)^10 = C_0 + C_1 y + C_2 y^2 + C_3 y^3+ cdots+C_{10}y^10#

so, deriving three times #(y + 4)^10# regarding #y# will result in

#10 xx 9 xx 8 xx(y+4)^7 = 3 xx 2 xx 1 xx C_3 + y xx p(y)#

Making now #y = 0# we get at

#C_3 = (10 xx 9 xx 8xx 4^7)/(3 xx 2 xx 1) = 1966080#