# 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

2n -8r +10 =0

+ -3n – 8r - 3 =0

-n = -7

n =7 and r = 3

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