$$\sum_{k=1}^n k(n-k+1) = \frac{1}{6}n(n+1)(n+2)$$
How do I prove this is true for all natural numbers n. I have done the basis and "n = k" steps but I cannot prove for "n = k+1".
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For clarity, the beforelast term should be written $(n-1)2$. – Oct 20 '20 at 11:55
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Don't write $...$ for something that goes down and then up. Use 2 $...$ instead. Also as the question is written right now it's not correct, unless I didnt understood what you meant in the $...$. – Tomer Wolberg Oct 20 '20 at 11:56
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@MauroALLEGRANZA; n=1 is odd – JMP Oct 20 '20 at 11:59
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@MauroALLEGRANZA; the parity of the number of terms is the same as the parity of n. – JMP Oct 20 '20 at 12:01
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Does this answer your question? induction proof: $\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$ – Arnaud D. Oct 20 '20 at 14:30
4 Answers
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By the assumption of the induction $$1(n+1)+2n+...+n\cdot2+(n+1)\cdot1=$$ $$=1(n+1)+2(n-1+1)+...+n(1+1)+n+1=$$ $$=\frac{1}{6}n(n+1)(n+2)+1+2+...+n+(n+1)=$$ $$=\frac{1}{6}n(n+1)(n+2)+\frac{(n+1)(n+2)}{2}=\frac{(n+1)(n+2)(n+3)}{6}.$$
Michael Rozenberg
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The sum is $$S_n=\sum_{k=1}^n k(n+1-k).$$
By comparison, we have $$S_{n+1}=\sum_{k=1}^{n+1} k(n+2-k)=\sum_{k=1}^{n} k(n+2-k)+n+1=S_n+\sum_{k=1}^nk+n+1 \\=S_n+\frac{(n+1)(n+2)}2.$$
On the other hand,
$$\frac{(n+1)(n+2)(n+3)}6-\frac{n(n+1)(n+2)}6=\frac{n+3-n}6(n+1)(n+2).$$
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The left-hand side is $\sum_{k=1}^nk(n+1-k)=(n+1)\sum_kk-\sum_kk^2$. Prove by induction$$\sum_kk=\tfrac12n(n+1),\,\sum_kk^2=\tfrac16n(n+1)(2n+1)$$so$$\sum_kk(n+1-k)=\tfrac16n(n+1)[3(n+1)-(2n+1)]=\tfrac16n(n+1)(n+2).$$Alternatively, use the hockey-stick identity.
J.G.
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If we increase $n$ to $n+1$, then the second term goes up by:
$$\frac{(n+1)(n+2)(n+3)}{6}-\frac{n(n+1)(n+2)}{6}=\frac{(n+1)(n+2)}{2}$$
and each of the first terms increase by $$k(n+1-k+1)-k(n-k+1)=k$$
and so the whole LHS increases by $(1+2+\dots+n) + (n+1)$, which is the same as the RHS.
JMP
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