2

I have to do a proof of the statement: $$N(b)\equiv2(2b+1)^2(2b+2)^2=\sum_{k=0}^{2b-1}\Biggl(\prod_{j=1}^4(4b-k+j)\biggl)+\prod_{j=1}^4(2b+j)-(2b+1)\prod_{j=1}^3(2b+j)$$ I know I'm not supposed to ask a question and then have someone solve it, but I really just need at least a place to start this proof. It looks like it needs to be done by induction, but it seems too complicated to substitute $N(b)$ into $N(b+1)$, or maybe I am wrong.

I have come to this statement by attempting to deductively prove that the alternating sum of every row (apart from the $0$th) of Pascal's triangle is equal to $0$. I know it is a lot easier to do this by induction, and I understand how to, but still wanted to know if you could do this by deduction which lead me to the statement above. This also leads to the possibility of error, so the statement above may not be true.

Here is my working out if it helps: Note that $m\in Z_{\ge0}$

Could someone please explain how to prove this? And possibly show how?

  • 2
    "every row of Pascal's triangle is equal to $0$"??? – Gary Jan 03 '22 at 04:56
  • @Gary my bad, I meant the alternating sum. Will edit –  Jan 03 '22 at 04:57
  • 1
    Is it supposed to be $$\left(\sum_{k=0}^{2b-1} \left(\prod_{j=1}^4(4b-k+j)\right)\right)+\left(\prod_{j=1}^4(2b+j)\right)-(2b+1)\left(\prod_{j=1}^3(2b+j)\right),$$ or is it something else? The nested summation and products are too ambiguous for me. – VTand Jan 03 '22 at 05:04
  • @VTand Yes, it is supposed to be like that. I will make a small edit. –  Jan 03 '22 at 05:07
  • 1
    Neither of your tags is approriate. You are not asking for an explanation of a given proof, nor for alternative proofs. Please make an effort to use correct tags. – Arturo Magidin Jan 03 '22 at 05:28

1 Answers1

1

The equality doesn't hold for some $b$. With $b = 1$, we have \begin{align*} & \left(\sum_{k=0}^{2b-1} \left(\prod_{j=1}^4(4b-k+j)\right)\right)+\left(\prod_{j=1}^4(2b+j)\right)-(2b+1)\left(\prod_{j=1}^3(2b+j)\right) \\ = \: & \left(\sum_{k=0}^{1} \left(\prod_{j=1}^4(4-k+j)\right)\right)+\left(\prod_{j=1}^4(2+j)\right)-3\left(\prod_{j=1}^3(2+j)\right) \\ = \: &\left(\prod_{j=1}^4(4-0+j) + \prod_{j=1}^4(4-1+j) \right) + (3 \cdot 4 \cdot 5 \cdot 6) - 3 \cdot (3 \cdot 4 \cdot 5) \\ = \: &\left(5 \cdot 6 \cdot 7 \cdot 8+ 4 \cdot 5 \cdot 6 \cdot 7\right) + (3 \cdot 4 \cdot 5 \cdot 6) - 3 \cdot (3 \cdot 4 \cdot 5) \\ = \: & 1680 + 840 + 360 - 180 \\ = \: & 2700, \end{align*} while $2(2b+1)^2(2b+2)^2 = 2 \cdot 3^2 \cdot 4^2 = 288$.

VTand
  • 2,634