# Thread: find kth term in a series

1. ## find kth term in a series

So i am trying to find the kth term for a series used to approximate (1+x)^n

the series is: 1+ n*x+ ((n(n-1))/2!)x^2 +((n(n-1)(n-2))/!)x^3 + ((n(n-1)(n-2)(n-3))/4!)x^4.....

i have figured out the denominatior but i cant figure out the numerator. any help? thanks.

2. Originally Posted by minnesotaeuro
So i am trying to find the kth term for a series used to approximate (1+x)^n

the series is: 1+ n*x+ ((n(n-1))/2!)x^2 +((n(n-1)(n-2))/!)x^3 + ((n(n-1)(n-2)(n-3))/4!)x^4.....

i have figured out the denominatior but i cant figure out the numerator. any help? thanks.
Well, yeah, the denominator is n! isn't it!

compare the numerators to the power of x: when
k= 1 numerator is n
k= 2 numerator is n(n-1)
k= 3 numerator is n(n-1)(n-2)
k= 4 numerator is n(n-1)(n-3)

Isn't it obvious that this is n(n-1)(n-2)...(n-(k-1))?

If you are thinking you need to write that as a "closed form", consider that
n(n-1)(n- 2)... (n-a)= [n(n-1)(n-2)...(n-a)(n-a-1)...(3)(2)(1)]/(n-a-1)(n-a-2)...(3)(2)(1)]= n!/(n-a-1)!