Combinatorics

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

#Combinatorics #HelpMe! #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Sadman Sakib - 7 years, 7 months ago

no, it was the alternating sum.

Sadman Fahim - 7 years, 7 months ago

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 - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$