Yes, there is a fast way to do it

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Find the sum of the squares of each number in the 34th row of Pascal's triangle.

Enter your answer as the sum of the digits of the number.

The answer is 39.

1 solution

Dallin Richards
Mar 16, 2016

Using the identity $\large \sum_{k=0}^{n} {n \choose k}^{2} = {2n \choose n}$ , we know that $\large \sum_{k=0}^{34} {34 \choose k}^{2} = {68 \choose 34}$ , which equals 28453041480000000000000000000, and the sum of these digits is $\boxed{39}$ .

Are we counting the top row as the 0th row? In any case we have $\binom{68}{34} = 28453041475240576740$ .

Chaitanya Rao - 1 year, 4 months ago

Yes, there is a fast way to do it

The answer is 39.

1 solution

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