Find #sum_(k=1)^ook x^k# ?

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Answer 1 · 2018-01-20T01:33:38

See below.

For #abs x < 1# we have

#sum_(k=0)^oo x^k = 1/(1-x)# and

#d^2/(dx^2)(1/(1-x) ) = 2/(1-x)^3 = sum_(k=2)^oo k(k-1)x^(k-2)#

then

#(x^2-x)/(1-x^3) = (x^2-x)/2sum_(k=2)^oo k(k-1)x^(k-2) = #

#= 1/2sum_(k=2)^oo k(k-1)x^k-1/2 sum_(k=2)^oo k(k-1)x^(k-1) = #

#=-x+1/2(sum_(k=2)^oo( (k-1)k-k(k+1))x^k) =#

#=-sum_(k=1)^ook x^k#