If in the expansion of (1+x)n,the coefficient of pth and qth terms are equal and the relation between p and q is p-q =n . Find the value of p2+q2. where p≠q.
Find the term independent of x in the expansion of (2x−1x)10.
We know that the cofficient of rth term in the expansion of (1+x)n is nCr−1.
Therefore, coefficients of pth and qth terms in the expansion of (1+x)n are nCp−1., respectively.
It is given that these coefficients are equal.
Also given,⇒p−q=n (ii)
On adding Eqs,(i) and (ii) we get
On putting p=n+1 in Eq.(i), we get
Let (r+1) th term be independent of x in the given expansion (2x−1x)10.
Then , Tr+1=10Cr(2x)10−r(−1x)r
⇒ Tr+1=(−1)r 10Cr(2)10−r(X)10−2r
For this term to be independent of x, we must have
So, (5+1) i.e. 6 th terms is independent of x, i.e T6=(−1)5 10C5(2)10−5