# Any shortcut?

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, 9 months ago

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me also waiting for this

- 2 years, 9 months ago

then please share it again and again

- 2 years, 8 months ago

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

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

- 2 years ago

- 2 years ago

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

- 2 years ago

okay you can write the formula here

- 2 years ago

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

- 2 years ago

- 2 years, 9 months ago

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!

- 1 year, 10 months ago

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

- 1 year, 10 months ago