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I encountered a problem which is asks for the first negative term in the expansion of $(1+x)^{\frac{27}{5}}$, where $x$ is always positive,the solution suggested in my module is like this:

$\displaystyle \text{ For the first negative term: } n-r+1 \lt 0 \text{ or, } r \gt 6\frac{2}{5}$, since $n = \frac{27}{5}$

Hence, the 8th term is the first negative term.

But my problem is that I am not getting the rationale behind this approach.Also,if you are aware of any other method for the same problem you may like to post too.

PS:Inquisitive to verify the same using Mathematica,I tried Expand[(1 + x)^(27/5)] but I guess this is not the correct function.

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up vote 2 down vote accepted

Hint: Use generalized binomial theorem, when does a negative factor appear in the binomial coefficient. (I am asssuming $x$ is positive here and $n,r$ have their usual meanings).

${n\choose r}=\frac{(n)(n-1)...(n-r+1)}{r!}$

Edit: $r,x$ are positive here. So the only way we can get a negative term is having a negative coefficient and hence only by having a negative numerator of the coefficient. Since the numerator is a product of real numbers it first becomes negative when the last factor is negative (since the first factor is positive and the value of the factors keeps decreasing by one successively). So this happens when $n-r+1<0$. Set $n=\frac{27}{5}$, so $\frac{27}{5}-r+1<0\implies \frac{32}{5}-r<0 \implies r>\frac{32}{5}=6\frac{2}{5}$

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Unfortunately, I still don't understand,could you explain. – Quixotic Nov 29 '10 at 5:57
@Debanjan: See edit – Timothy Wagner Nov 29 '10 at 6:02
+1,Now as $r>32/5$ we can write $r \approx 7$, hence the 8th term is the answer.-- correct? – Quixotic Nov 29 '10 at 6:11
@Debanjan: Precisely, as $r$ starts with $0$. – Timothy Wagner Nov 29 '10 at 6:12
Yes,thank you very much. – Quixotic Nov 29 '10 at 6:13

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