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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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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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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!

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

no, it was the alternating sum.

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestDo 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!

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The sum of the numbers in row

\(n\)is\(2^{n-1}\)….simple .Log in to reply

no, it was the alternating sum.

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