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What is the sum of the second numbers in the first $100$ rows of Pascal's triangle (excluding the first row, the row containing a single $1$)? The sum should be from the second to the hundredth row.

Pascal's triangle

Starting from the second row, I initially thought this meant you count from the left two numbers. So it would be $$1+2+3+4+\cdots+99$$ This means I get $4950$. I thought this would be too simple of a solution.

Could someone tell me if the addition I did above is all the question is asking from me?

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    Except that perhaps the question wants you to use rows 2 through 101 (excluding the first row from the count of rows), that seems like a reasonable interpretation. – rogerl Aug 30 '19 at 15:22
  • @rogerl, the question as it appears now specifically says to add up $99$ numbers (from “the second to hundredth row”), not $100$. In other words, “the first 100 rows ... excluding the first row” is meant to specify $99$ rows, as opposed to what I think you surmised. (So it’s the “first 100 ... excluding,” and not “first 100 ... commencing with.”) – Steve Kass Aug 30 '19 at 21:26
  • Then I think your answer is correct. – rogerl Aug 30 '19 at 22:03

1 Answers1

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To make the computations more transparent, I'll start indexing with 0. Therefore we want to sum up the elements $a_{k,1}$ ($k\geq 1$). I'll also assume, that the elements, that are 'outside' of the triangle are all equal 0 (in particular $a_{k,k+1}=0$ and $a_{0,1}=a_{0,2}=a_{1,2}=0$)

Note, that for $n\geq 1$ we have: $$a_{n,2}=a_{n-1,1}+a_{n-1,2}$$ Therefore by consecutive replacing the element $a_{*,2}$ on the right side with analogical value, we obtain $$a_{n,2}=a_{n-1,1}+a_{n-2,1}+...+a_{1,1}+a_{0,1}$$ We have then the sum of $n$ elements with he index 1 from the first $n$ rows.

Of course $a_{i,j}=\binom{i}{j}$

For $n=100$ we have then: $$\sum_{k=0}^{99}a_{k,1}=a_{100,2}=\binom{100}{2}$$

Jaroslaw Matlak
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