The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r.

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

Here the binomial is (1+x)n with a = 1 , b = x and n = n


The (r+1)th term is given by


T(r+1) = nCr 1n-r xr


T(r+1) = nCr xr


The coefficient of (r+1)th term is nCr


The rth term is given by (r-1)th term


T(r+1-1) = nCr-1 xr-1


Tr = nCr-1 xr-1


the coefficient of rth term is nCr-1


For (r-1)th term we will take (r-2)th term


Tr-2+1 = nCr-2 xr-2


Tr-1 = nCr-2 xr-2


the coefficient of (r-1)th term is nCr-2


Given that the coefficient of (r-1)th, rth and r+1th term are in ratio


1:3:5









3r - 3 = n – r + 2


n - 4r + 5 =0…………1


Also









5r = 3n - 3r + 3


8r 3n - 3 =0………….2


We have 1 and 2 as


n - 4r 5 =0…………1


8r – 3n - 3 =0…………….2


Multiplying equation 1 by number 2


2n -8r +10 =0……………….3


Adding equation 2 and 3


2n -8r +10 =0


+ -3n – 8r - 3 =0


-n = -7


n =7


and r = 3


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