- #1
Reingley
- 8
- 0
Working through Leonard Susskind's book The Theoretical Minimum, I noticed an issue with his expansion for the Binomial Expansion (he was missing factorials in the denominators). This led me to some confusion about the final term that is generally written (bn).
(a+b)n = an + nan-1b + n(n-1)/2! an-2b2 + ... + bn
My issue with this is that if one were to solve for n = 2, the b2 term comes out from the 3rd term (2! term) and there is no need to add the bn=2 term at the end.
Is my problem that if I am using n = 2, I shouldn't even bother to include the 3rd term (and all higher order terms that incidentally go to zero anyways)? It seems to me that the bn term is simply unnecessary at the end.
Thanks for any input!
(a+b)n = an + nan-1b + n(n-1)/2! an-2b2 + ... + bn
My issue with this is that if one were to solve for n = 2, the b2 term comes out from the 3rd term (2! term) and there is no need to add the bn=2 term at the end.
Is my problem that if I am using n = 2, I shouldn't even bother to include the 3rd term (and all higher order terms that incidentally go to zero anyways)? It seems to me that the bn term is simply unnecessary at the end.
Thanks for any input!