Is there a formula to get r+2r2+3r3+⋯+nrn provided that |r|<1? This seems like the geometric "sum" r+r2+⋯+rn so I guess that we have to use some kind of trick to get it, but I cannot think of a single one. Can you please help me with this problem?
Answer
We have
1+r+r2+⋯+rn=rn+1−1r−1
Differentiating this once, we obtain
1+2r+3r2+⋯+nrn−1=nrn+1−(n+1)rn+1(r−1)2
Multiply the above by r to obtain what you want.
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