New user? Sign up

Existing user? Sign in

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, 7 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

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

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

me also waiting for this

Log in to reply

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 .

Problem Loading...

Note Loading...

Set Loading...

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

Sort by:

TopNewestme also waiting for this

Log in to reply

then please share it again and again

Log in to reply

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!

Log in to reply

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

Log in to reply