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