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# how do you solve (x+1)^100?

This question has been bugging me all weekend=s

I'm not really sure how to be more specific about it. This is all i was given. How would you expand it? I looked into the binomial theorem and pascals triangle, but i don't seem to understand either that well. Thanks though

### 6 Answers

- 1 decade agoFavorite Answer
If you want to expand it, you can write it as Sum[(100,k) x^k, k, 0, 100], a sum from 0 to 100 over the index k. The coefficients (100,k) are called binomial coefficients, and they have the value 100!/[k! (100-k)!] (read about factorials if you don't know this notation). Some of the terms look like:

(x+1)^100 =

(100,100) x^100 + (100,99) x^99 + (100,98) x^98 + ... + (100,1) x + (100,0) =

100!/[100! (100-100)!] x^100 + 100!/[99! (100-99)!] x^99 + 100!/[98! (100-98)!] x^98 + .. + 100!/[1! (100-1)!] x + 100!/[0! (100-0)!] =

100!/[100! 0!] x^100 + 100!/[99! 1!] x^99 + 100!/[98! 2!] x^98 + ... + 100!/[1! 99!] x + 100!/[0! 100!] =

1/0! x^100 + 100/1! x^99 + 100*99/2! x^98 + ... + 100/1! x + 1/0! =

x^100 + 100 x^99 + 4950 x^98 + ... + 100 x + 1

- 1 decade ago
(x+1)^100 equals what? Or do you mean if x = 2 then your term equals 3^100?

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- Anonymous1 decade ago
use binomial theorem.

- Anonymous1 decade ago
What is x equal?