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  • If 0 < x < 1, then the first negative term in the expansion of <img alt="(1+x)^{\frac{27}{5}}" src="http://entrancecorner.oncodecogs.com/gif.latex?%281&plus;x%29%5E%7B%5Cfrac%7B

If 0 < x < 1, then the first negative term in the expansion of (1+x)^{\frac{27}{5}} is

Option: 1

8th term


Option: 2

7th term


Option: 3

6th term


Option: 4

9th term


Answers (1)

best_answer

Binomial Theorem for any  Index

\\\text{For negative or fractional Index and } |x|<1, \\\\\mathrm{(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+......+\frac{n(n-1)(n-2).....(n-r+1)}{r!}x^r....\infty}

 

Now,

First term is 1, so positive

Second term is also positive as both n and x are positive 

In any term, xr is positive, r! is positive. So, the factor that will make a term negative is (n - r + 1)

So, we need to find r when (n - r + 1) will be negative for the first time (r is an integer), where n=27/5

Solving 27/5 - r + 1 < 0, we get r > 6.4 

So, r = 7 and this happens in 8th term, so 8th term is the answer.

Option A is correct.

Posted by

Devendra Khairwa

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