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प्रश्न
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
उत्तर
In (a + x)n general term is tr + 1 = nCr
So, the coefficient of tr + 1 is nCr
We are given that the coefficients of three consecutive terms are in the ratio 1 : 7 : 42.
⇒ nCr –1 : nCr : nCr + 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
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