Question
Find the 6th term in the expansion (x^2 - 1/x)^25
It is a Binomial Theorem problem and I am extremely confused. Please and thank you!
It is a Binomial Theorem problem and I am extremely confused. Please and thank you!
Answers
Shenaya
So this expansion has 26 terms.Let consider the (r+1)th term and let's name it T(r+1)
So T(r+1)= nCr*[(x^2)(n-r)]*[(1/x)^r
nCr=[n!]/[(n-r)!*(r)!]
= nCr*[(x^2)(n-r)]*[(x^-1)^r] = nCr*[(x)^(2n-2r-r)]=nCr*[(x)^(2n-3r)]
When r=5 ,we get T(6) which is the sixth term of the expansion.
So T(r+1)= nCr*[(x^2)(n-r)]*[(1/x)^r
nCr=[n!]/[(n-r)!*(r)!]
= nCr*[(x^2)(n-r)]*[(x^-1)^r] = nCr*[(x)^(2n-2r-r)]=nCr*[(x)^(2n-3r)]
When r=5 ,we get T(6) which is the sixth term of the expansion.