Question

# 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. Or Find the term independent of x in the expansion of (2x−1x)10.

Solution

## 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. ∴nCp−1.=nCq−1 ⇒p−a+q−1=n            [∵nCx=nCy⇒x+y=n] ⇒P+q=n+2                  (i) Also given,⇒p−q=n              (ii) On adding Eqs,(i) and (ii) we get          2p=2n+2⇒p=n+1 On putting p=n+1 in Eq.(i), we get                  q=n+2−(n+1)=1 Now, p2+q2=(N+1)2+12 =n2+1+2n+1=n2+2n+2 Or 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 10−2r=0⇒r=5 So, (5+1) i.e. 6 th terms is independent of x, i.e T6=(−1)5 10C5(2)10−5 =−10!5!5!×25=−252×32=−8064 Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More