Question

If the binomial coefficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42, then find n

Answer 1

In (a + x)ⁿ general term is t_{r + 1} = ⁿC_r

So, the coefficient of t_{r + 1} is ⁿC_r

We are given that the coefficients of three consecutive terms are in the ratio 1 : 7 : 42.

⇒ ⁿC_{r –1} : ⁿC_r : ⁿC_{r + 1} = 1 : 7 : 42

(i.e.) `(""^"n""C"_("r" - 1))/(""^"n""C"_"r") = 1/7`  ......(1)

and `(""^"n""C"_"r")/(""^"n""C"_("r" + 1)) = 7/42 = 1/6` ......(2)

(1) ⇒ `(("n"!)/(("r" - 1)!("n" - "r" - 1)!))/(("n"!)/("r"!("n" - "r")!)) = 1/7`

(i.e.) `("n"!)/(("r" - 1)!("n" + 1 - "r")!) xx ("r"!("n" - "r")!)/("n"!) = 1/7`

⇒ `"r"/("n" + 1 - "r") = 1/7`

⇒ 7r = n + 1 – r

⇒ 8r – n = 1 → (A)

(2) ⇒ `(("n"!)/("r"!("n" - "r")!))/(("n"!)/(("r" + 1)!("n" - "r" + 1)!["n" - "r" - 1])) = 1/6`

(i.e.) `("n"!)/("r"!("n" - "r")!) xx (("r" + 1)!("n" - "r" - 1)!)/("n"!) = 1/6`

`(("r" + 1))/("n" - "r") =1/6`

n – r = 6r + 6

n – 7r = 6 → (B)

Solving (A) and (B)
– n + 8r = 1  → (A)
   n – 7r = 6  → (B) 
(A) + (B) ⇒ r = 7

Substitting r = 7 in (B)

n = 6 + 7 × 7

n = 6 + 49 = 55