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In the binomial expansion of (a - b)^(n)...

In the binomial expansion of `(a - b)^(n) , n ge 5` , the sum of
the ` 5^(th) and 6^(th)` terms is zero. Then, `a//b` equals

A

`(n-5)/(6)`

B

`(n- 4) /(5)`

C

`(5)/(n-4)`

D

`(6)/(n-5)`

Text Solution

Verified by Experts

The correct Answer is:
B
We have,
`T_(5) + T_(6) = 0`
`rArr ""^(n)C _(4) ""a^(n - 4) b^(4) - ""^(n)C_(5) a^(n - 5) b^(5) = 0`
`rArr ""^(n)C_(4) = ""^(n)C_(5) b rArr (a)/(b) = (""^(n)C_(5))/(""^(n)C_(5)) rArr (a)/(b) = (n - 4)/(5)` .
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