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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!

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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

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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.

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I already asked it on mathstack

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

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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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