Use the trigonometric identity:
#sin^2(4x) = (1-cos(8x))/2#
We know the MacLaurin series of #cost#:
#cost = sum_(n=0)^oo (-1)^nt^(2n)/((2n)!)#
Substituting #t=8x# we have:
#cos 8x = sum_(n=0)^oo (-1)^n(8x)^(2n)/((2n)!) = sum_(n=0)^oo (-1)^n2^(6n)x^(2n)/((2n)!)#
and then:
#sin^2(4x) = 1/2 (1-sum_(n=0)^oo (-1)^n2^(6n)x^(2n)/((2n)!))#
Extract the term for #n=0# from the sum:
#sin^2(4x) = 1/2 (1-1 - sum_(n=1)^oo (-1)^n2^(6n)x^(2n)/((2n)!))#
and then bring the common factor #-1/2# inside the sum:
#sin^2(4x) = sum_(n=1)^oo (-1)^(n+1)2^(6n-1)x^(2n)/((2n)!)#
We can also change the index with #k=n-1# to have:
#sin^2(4x) = sum_(k=0)^oo (-1)^(k+2) 2^(6(k+1)) x^(2(k+1))/((2(k+1))!)#
#sin^2(4x) = 64x^2 sum_(k=0)^oo (-1)^k 2^(6k) x^(2k)/((2k+2)!)#
