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Combinatorics

What is the alternating sum of the numbers in row n of Pascal's triangle?

Note by Fatin Farhan
3 years, 9 months ago

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Do you mean \(\sum_{k=0}^n \binom{n}{k} (-1)^k\)? By the Binomial Theorem, \(\sum_{k=0}^n \binom{n}{k} x^k = (1+x)^n\), so we can just plug in \(x=-1\) and get the answer 0. Sanity check: 1-1=0, 1-2+1=0, 1-3+3-1=0. Not insane! Yay! Eric Edwards · 3 years, 9 months ago

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The sum of the numbers in row \(n\) is \(2^{n-1}\) ….simple . Sadman Sakib · 3 years, 9 months ago

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@Sadman Sakib no, it was the alternating sum. Sadman Fahim · 3 years, 9 months ago

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