If the coefficients of three consecutive terms in the expansion of (1+x)n  are 28,56  and 70,  then the value of n∈N  is

Given expansion of (1+x)n   If the consecutive terms are Tr,Tr+1  and Tr+2,  then we are given that general term Tr+1=nCr xn−r ar     Tr=T(r−1)=nCr−1 (1)n−r+1 xr−1=28     ⇒nCr−1=28               ….(1) ( coefficients of expansion)         Tr+1=T(r+1)=nCr (1)n−r xr=56         ⇒nCr=56              …..(2) ( coefficients of expansion)      And      Tr+2=T(r+1)+1=nCr+1 (1)n−r−1 xr+1=70   ⇒nCr+1=70              …. (3) ( coefficients of expansion)Dividing (1) by (2), we get  nCr−1nCr=2856⇒rn−r+1=12(∴nCr−1nCr=rn−r+1) And dividing (2) by (3), we get nCrnCr+1=5670⇒r+1n−r=45(∴nCrnCr+1=r+1n−r) r+1n−r=45 From here, we get n=2r+r−1 And  n=9r+54 ⇒  3r  −1  =9r+54 ⇒  r=3  and n=8