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(2) ∑<n=1,∞>n^2 x^n = ∑<n=1,∞>[(n+2)(n+1) - 3(n+1) +1] x^n
= [∑<n=1,∞> x^(n+2)]'' - 3 [∑<n=1,∞> x^(n+1)]' + ∑<n=1,∞>x^n
= [x^3/(1-x)]'' - 3[x^2/(1-x)]' + x/(1-x)
= [-x^2-x-1 + 1/(1-x)]'' - 3[-x-1 + 1/(1-x)]' + x/(1-x)
= -2 + 2/(1-x)^3 - 3[-1+1/(1-x)^2] - 1 + 1/(1-x)
= 2/(1-x)^3 - 3/(1-x)^2 + 1/(1-x)
= [2 -3(1-x) + (1-x)^2]/(1-x)^3 = x(1+x)/(1-x)^3, |x| < 1.
= [∑<n=1,∞> x^(n+2)]'' - 3 [∑<n=1,∞> x^(n+1)]' + ∑<n=1,∞>x^n
= [x^3/(1-x)]'' - 3[x^2/(1-x)]' + x/(1-x)
= [-x^2-x-1 + 1/(1-x)]'' - 3[-x-1 + 1/(1-x)]' + x/(1-x)
= -2 + 2/(1-x)^3 - 3[-1+1/(1-x)^2] - 1 + 1/(1-x)
= 2/(1-x)^3 - 3/(1-x)^2 + 1/(1-x)
= [2 -3(1-x) + (1-x)^2]/(1-x)^3 = x(1+x)/(1-x)^3, |x| < 1.
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