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?

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?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestme also waiting for this – Anshu Jha · 1 year, 5 months ago

Log in to reply

– Aman Rajput · 1 year, 5 months ago

then please share it again and againLog 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. – Pi Han Goh · 8 months, 2 weeks ago

Log in to reply

– Aman Rajput · 8 months, 2 weeks ago

I already asked it on mathstackLog in to reply

– Pi Han Goh · 8 months, 2 weeks ago

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

– Aman Rajput · 8 months, 2 weeks ago

okay you can write the formula hereLog in to reply

– Pi Han Goh · 8 months, 2 weeks ago

Look at the solutions in the link that I've given. That's the general approach.Log in to reply

@Calvin Lin sir ,,, will you please share this... i eagerly want to know this.... :/ – Aman Rajput · 1 year, 6 months ago

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! – Samanvay Vajpayee · 6 months, 3 weeks ago

Log in to reply

– Aman Rajput · 6 months, 3 weeks ago

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