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Prove by induction :

$$\displaystyle \sum _{k=0}^{n}( k+1)( n-k+1) =\binom{n+3}{3}$$

induction basis: $n=0$

$$\displaystyle \sum _{k=0}^{n}( 0+1)( 0-0+1) =1=\binom{3}{3}$$

For $n+1$:

\begin{aligned} \sum _{k=0}^{n+1}( n+1+1)( n+1-(n+1)+1) & =\sum _{k=0}^{n}( k+1)( n-k+1) +( n+2)(( n+1) -( n+1) +1)\\ & =\binom{n+3}{3} +( n+2)\\ \end{aligned}

I had to stop here because I realize there is a mistake ...

Unfortunately, I didn't succeed in many ways.

Martin Sleziak
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ATB
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    I know you asked for an induction proof so this doesn't answer that aspect, but a combinatorial proof is readily available for this. On the RHS this counts the number of subsets of size $3$ of the set ${-1,0,1,2,\dots,n,n+1}$ while the LHS counts the same but by breaking apart based on the value of the middle element, calling it $k$, noting that if the middle value was $k$ there are $k+1$ choices smaller than $k$ remaining in ${-1,0,1,\dots,n,n+1}$ for the smallest number in the subset and there are $(n-k+1)$ choices larger. – JMoravitz May 03 '21 at 15:14
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    Hint: $(n+2)^2= (k+1 + n-k+1)^2 = (k+1)^2 + 2 (k+1)(n-k+1) + (n-k+1)^2$ – Gribouillis May 03 '21 at 15:52
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    @JMoravitz Yep: https://math.stackexchange.com/a/4124710/683666 – RobPratt May 03 '21 at 16:16
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    This reminded me of a question I have asked some time before: How to show that $\sum_{k=1}^n k(n+1-k)=\binom{n+2}3$?. You can see various related posts among the linked questions. – Martin Sleziak May 22 '21 at 09:37

1 Answers1

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You set up the induction correctly by getting the base case and stating your induction hypothesis.

For the induction step:

$\sum\limits_{k=0}^{n+1}(k+1)((n\color{red}{+1})-k+1)$ You seem to have forgotten this red $\color{red}{+1}$.

So, we have $\sum\limits_{k=0}^{n+1}(k+1)((n\color{red}{+1})-k+1) = \sum\limits_{k=0}^n(k+1)(n-k+1)\color{red}{+\sum\limits_{k=0}^n(k+1)}+(n+2)$.

This should hopefully get you back on track.

The first summation simplifies to $\binom{n+3}{3}$ per induction hypothesis, the second summation simplifies to $\binom{n+2}{2}$ recognizing it as the $n+1$'st triangular number. So we have $\binom{n+3}{3}+\binom{n+2}{2}+\binom{n+2}{1}=\binom{n+3}{3}+\binom{n+3}{2}=\binom{n+4}{3}=\binom{(n+1)+3}{3}$ using Paschal's identity to finish.

JMoravitz
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  • Thank you, I succeeded with the induction, but I didn't get why we are adding the sigma of k+1, where the sigma k+1 came from? just if you can explain the sigma after the original sigma in the line I will appreciate that. – ATB May 03 '21 at 16:37
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    @ATB Do you see where the red $+1$ came from in the initial set up? Do you see why $\sum a(b+1)=\sum ab + \sum a$? Now, take as analogy rather than $a$ have $(k+1)$ and instead of $b$ we had $(n-k+1)$ and $b+1$ as $((n+1)-k+1)$. – JMoravitz May 03 '21 at 17:04
  • Thanks, Understood, and where the last element $(n+2)$ came from? – ATB May 04 '21 at 07:36
  • @atb from breaking off the last term in the summation to be handled separately – JMoravitz May 04 '21 at 11:32