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# Interesting Maths 2008

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- JacobLv 61 decade agoFavorite Answer
First is to find Σ(k=1 to ∞) k^3*r^k (for |r|<1)

Since Σ(k=1 to ∞) r^k=r/(1-r) for |r|<1

diff. both side term-by-term w.r.t. r,

Σ(k=1 to ∞) kr^(k-1)=1/(1-r)^2

multiply r both side and we get

Σ(k=1 to ∞) kr^k=r/(1-r)^2

repeat the same process, we finally get

Σ(k=1 to ∞) k^3 r^k==r(1+4r+r^2)/(1-r)^4

putting r=1/3<1, (the series still converges)

Σ(k=1 to ∞) k^3/3^k=33/8

Source(s): ME

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