Question

if the coefficient of (r-1)th, rth and (r+1)th terms in the expansion of (1+x)^n are in the ratio 1:7:42, find n and r

Answer 1

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! / (r-2)! (n -r +2)! ] x [(r-1)!(n-r+1)!/ n!] = 1/3

( r -1)(r -2)! / ( r – 2)! ( n -r +2) = 1 / 3
( r -1) / ( n -r +2) = 1 / 3

3r - 3 = n - r + 2

n - 4r = -5 ----------(1)

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

Answer 2

Answer:

n=7

r=2

Step-by-step explanation:

r+1/n-r=3/5

r/n-r+1=1/3

5r+5=3n-3r

==8r=3n-5

==3r=n-r+1

n=7,,,r=2