Question

If $ x $ positive, the first negative term in the expansion of $ {(1 + x)^{27/5}} $ is:
(A) seventh term
(B) fifth term
(C) eight term
(D) sixth term

Answer 1

Hint: Use the expansion formula of binomial theorem. Then check for the condition that a coefficient of $ {x^ + } $ needs to satisfy to be negative and solve by applying that condition to the above expression.

Complete step-by-step answer:
We have a formula for $ {(1 + x)^n} $ as
 $ {(1 + x)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + ........... $
Where, $ {}^nCr = \dfrac{{n!}}{{N!(n - r)!}} $
Thus, $ {(1 + x)^n} $ can also be written as
 $\Rightarrow {(1 + x)^n} = \dfrac{{n!}}{{0!(n - 0)!}} + \dfrac{{n!}}{{1!(n - 1)!}}x + \dfrac{{n!}}{{2!(n - 2)!}}{x^2} + ........ $
Since, we know that $ 0! = 1! = 1, $ and $ n! = n(n - 1)! $
 $ = 1 + \dfrac{{n(n - 1)!}}{{(n - 1)!}}x + \dfrac{{n(n - 1)(n - 2)!}}{{2!(n - 2)!}}{x^2} + ....... $
By simplifying it, we get
 $\Rightarrow {(1 + x)^n} = 1 + nx + \dfrac{{n(n - 1)}}{{2!}}{x^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + .... $
By using above formula, we can write
 $\Rightarrow {(1 + x)^{27/5}} = 1 + \dfrac{{2x}}{5} + \dfrac{{\dfrac{{27}}{5}\left( {\dfrac{{27}}{5} - 1} \right)\left( {\dfrac{{27}}{5} - 2} \right)}}{{2!}}{x^2} + \dfrac{{\dfrac{{27}}{5}\left( {\dfrac{{27}}{5} - 1} \right)\left( {\dfrac{{27}}{5} - 2} \right)}}{{3!}}{x^3}....... $
If we observe the above equation carefully, we can say that the above expansion will have the first negative term, when it has the term.
 $ \dfrac{{\dfrac{{27}}{5}\left( {\dfrac{{27}}{5} - 1} \right)\left( {\dfrac{{27}}{5} - 2} \right)...\left( {\dfrac{{27}}{5} - r} \right)}}{{r!}}{x^{r + 1}} $
Such that $ \dfrac{{27}}{5} - r $ is negative
i.e., we have smallest $ r $ such that
 $ \dfrac{{27}}{5} - r < 0 $
 $ \Rightarrow \dfrac{{27}}{5} < r $
 $ \Rightarrow r > \dfrac{{27}}{5} $
 $ = 5.4 $
Therefore, the smallest integer $ r $ that is greater than $ 5.4 $ is $ r = 6. $
Hence, sixth term will be the first negative term in the expansion of $ {(1 + x)^{27/5}} $
Therefore, from the above explanation the correct option is (D) sixth term.
So, the correct answer is “Option D”.

Note: You cannot keep on simplifying every coefficient of $ x,{x^2},{x^3} $ etc.
To check for the negative term. That would be too time consuming. You have to think smart to solve such questions.
The first negative term would come only when some coefficient becomes negative.
And that would happen only if $ \left( {\dfrac{{27}}{5} - r} \right) $ is negative for some $ r. $