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Coefficient of \(x^{49}\) in expansion of \((x+1)(x+2)(x+3)\dots(x+100)\)

or you can generalise it for any power coefficient?

Note by Aman Rajput 2 years, 3 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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me also waiting for this

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then please share it again and again

Generalize? Yes, but the formula is too long and worth writing it down.

To solve this question, look at a similar question here.

I already asked it on mathstack

I'm pessimistic that the formula will be helpful. It's like trying to remember all 4 quartic formulas.

@Pi Han Goh – okay you can write the formula here

@Aman Rajput – Look at the solutions in the link that I've given. That's the general approach.

@Calvin Lin sir ,,, will you please share this... i eagerly want to know this.... :/

Very simple . Use vietta's to generalise . Like : let f(x)= (x+1)(x+2)(x+3)......(x+n) then , f(x)= x^n + (1+2+3+...n).x^(n-1) + (1.2+1.3+1.4+...1.n+ 2.3+2.4+...2.n+3.4+3.5 +...).x^(n-2) +... I hope it is clear now!

so tell me the answer of the my question... i know that what you have written .

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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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TopNewestme also waiting for this

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then please share it again and again

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Generalize? Yes, but the formula is too long and worth writing it down.

To solve this question, look at a similar question here.

Log in to reply

I already asked it on mathstack

Log in to reply

I'm pessimistic that the formula will be helpful. It's like trying to remember all 4 quartic formulas.

Log in to reply

Log in to reply

Log in to reply

@Calvin Lin sir ,,, will you please share this... i eagerly want to know this.... :/

Log in to reply

Very simple . Use vietta's to generalise . Like : let f(x)= (x+1)(x+2)(x+3)......(x+n) then , f(x)= x^n + (1+2+3+...n).x^(n-1) + (1.2+1.3+1.4+...1.n+ 2.3+2.4+...2.n+3.4+3.5 +...).x^(n-2) +... I hope it is clear now!

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so tell me the answer of the my question... i know that what you have written .

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