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Question

The coefficients of the (r1)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.

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Solution

Given coefficients of( r-1)th, rth and (r+1)th terms in the expansion of (x+1)n are in the ratio 1:3:5
Coefficient of (r-1)th term = C(n, r-2)
Coefficient of rth term = C(n, r-1)
Coefficient of (r+1)th term = C(n, r)
C(n, r-2) : C(n, r-1) : C(n, r) =
1:3:5
Divide 1st and 2nd

C(n,r-2)/C(n,r-1) = 1 / 3

= [n!(r2)!(nr+2)!][(r1)!(nr+1)!n!] = 1/3


[(r1)!(r2)!(r2)!(nr+2)!] = [13]

C(n,r-1)/C(n,r) = 3/5( r -1) / ( n -r +2) = 1 / 3
3r - 3 = n - r + 2
n - 4r = -5 ----------(1)
Divide 2nd and 3rd

[n! / (r-1)! (n -r +1)! ] x r!(n-r)!/ n! = 3 / 5

r / (n -r + 1) = 3 / 5

5r = 3n - 3r +3
3n - 8r = -3 ---------(2)
2(n - 4r = -5)
2n - 8r = -10 ---------(3)
Subtract (3) from (2)
n = 7
Substitute n = 7 in (2)
We get r = 3
n = 7, r = 3


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