# Combinatorics

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

Note by Fatin Farhan
4 years, 8 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!

- 4 years, 8 months ago

The sum of the numbers in row $$n$$ is $$2^{n-1}$$ ….simple .

- 4 years, 8 months ago

no, it was the alternating sum.

- 4 years, 8 months ago