Public Administration Laxmikant Pdf Download ((HOT))
Public Administration Laxmikant Pdf Download
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Proving inductively $\sum_{k=1}^n (k+1)^m = 2\sum_{k=1}^{n-1} k^m$
I am trying to prove inductively that $\sum_{k=1}^n (k+1)^m = 2\sum_{k=1}^{n-1} k^m$. In general, if $\sum_{k=1}^n f(k) = \sum_{k=1}^{n-1} f(k)$ holds for $n-1$, then $\sum_{k=1}^n f(k) = 2 \sum_{k=1}^{n-2} f(k)$ holds for $n$.
I started the problem with $n = 1$, and the second case can be proven with $m = 1$ (this is the same formula as inductive step for $m = 1$).
For the inductive step I do not know how to proceed. I tried to prove the case $m = 2$, but I am stuck at that point. I can’t seem to get anywhere from it. I will be happy with hints.
A:
You should start with the simpler version:
$$
\sum_{k=1}^{n}k^m=\frac{n(n+1)}2\sum_{k=0}^{n-
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