If coefficients of three successive terms in the expansion of (x+1)n  are in the ratio 1:3:5 , then n  is equal to

If the three consecutive terms be Tr,Tr+1,Tr+2,  then we are given that nCr−1:nCr:nCr+1::1:3:5 . From here, we Get nCr−1nCr=13 ⇒  n!(n−r+1)!(r−1)!.(n−r)!r!n!=13 ⇒  n!(n−r)! (n−r+1) (r−1)!.(n−r)! (r−1)! rn!=13 (∴ r!=(r−1)! r) ⇒  rn−r+1=13⇒n−r+1=3r⇒n−4r+1=0  ................(1)                         And     nCrnCr+1=35 ⇒  n!(n−r)!r!.(n−r−1)!(r+1)!n!=35  ⇒  n!(n−r)(n−r−1)!r!.(n−r−1)!(r+1)r!n!=35 (∴(r+1)!=(r+1)r!) ⇒  r+1n−r=35⇒5r+5=3n−3r⇒3n−8r−5=0.............(2)                           …(2) Solving Eq(1) and Eq(2), we get n=7,r=2