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hmph
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Homework Statement
I am trying to prove how \(g''(r)=\sum\limits_{k=2}^\infty ak(k-1)r^{k-2}=0+0+2a+6ar+\cdots=\dfrac{2a}{(1-r)^3}=2a(1-r)^{-3}\).
I don't know what I am doing wrong and am at my wits end.
The Attempt at a Solution
(The index of the summation is always k=2 to infinity)Ʃak(k-1)r^(k-1)
=a Ʃ k(k-1)r^(k-2)
=a Ʃ (r^k)''
=a (r^2 / (1-r)''
From this point I get a mess, and the incorrect answer. The thing I have a problem is I think Ʃ (r^k)'' when k is from 2 to infinity is r^2/(1-r), since the first term in this sequence is r^2. I don't think that is correct though