NCERT Solutions
Class 11 Maths
Binomial Theorem
Ex.8.2 Q.10
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.
It is known that (k + 1) th term, (Tk+1), in the binomial expansion of (a + b) n is given by
Tk+1 = nCk an-k bk
Therefore, (r – 1) th term in the expansion of (x + 1) n is
Tr-1 = nCr-2 (x)n-(r - 2) 1r-2 = nCr-2 (x)n - r + 2
(r + 1) term in the expansion of (x + 1) n is
Tr+1 = nCr (x)n-r 1r = nCr (x)n - r
rth term in the expansion of (x + 1) n is
Tr = nCr-1 (x)n-(r - 1) 1r-1 = nCr-1 (x)n - r + 1
Therefore, the coefficients of the (r – 1) th, rth, and (r + 1) th terms in the expansion of (x + 1) n are
respectively. Since these coefficients are in the ratio 1:3:5, we obtain
(nCr-2 ÷ nCr-1) = 
and (nCr-1 ÷ nCr) = 
Now, nCr-2 ÷ nCr-1 =
=> [n! ÷ {(r - 2)! × (n – r + 2)!}] ÷ [{n! ÷ {(r - 1)!} × (n – r + 1)!}] =
=> [(r - 1) × (r - 2)! × (n – r + 1)!] ÷ [(r - 2) × (n - r + 2)! × (n – r + 1)!] =
=> (r - 1) ÷ (n – r + 2) =
=> 3(r - 1) = (n – r + 2)
=> 3r – 3 = n – r + 2
=> n – 4r + 5 = 0 ………… (1)
Again, nCr-1 ÷ nCr =
=> [n! ÷ {(r - 1)! × (n – r + 1)!}] ÷ [n! ÷ {r! × (n – r)!}] =
=> [r × (r - 1)! × (n - r)!] ÷ [(r - 1) × (n - r + 1)! × (n – r)!] =
=> r ÷ (n – r + 1) =
=> 5r = 3(n – r + 1)
=> 5r = 3n – 3r + 3
=> 3n – 8r + 3 = 0 ………… (2)
Solving equations (1) and (2), we get
Multiplying equation 1 by 3 and subtracting it from equation (2), we obtain
4r – 12 = 0
=> r = 3
Putting the value of r in equation (1), we obtain
n – 12 + 5 = 0
=> n = 7
Thus, the values of n and r are 7 and 3 respectively.