Key Concept: Binomial Theorem for Fractional and Negative Indices

For any real number $n$ (including fractions or negative integers) and for $|x| < 1$, the binomial expansion of $(1+x)^n$ is given by an infinite series. The general term, denoted as $T_{r+1}$ (which is the $(r+1)^{th}$ term), is expressed as: $T_{r+1} = \binom{n}{r} x^r$ where the binomial coefficient $\binom{n}{r}$ is defined as: $\binom{n}{r} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!}$ Here, $r$ represents the power of $x$ in the term.

Step-by-step Derivation

1. Identify Given Values and Expression: The given expression is ${\left( {1 + x} \right)^{27/5}}$. Comparing this with the general form $(1+x)^n$, we identify $n = \frac{27}{5}$. We are also told that $x$ is positive ($x > 0$).

2. Analyze the Sign of the General Term ($T_{r+1}$): The general term is $T_{r+1} = \binom{n}{r} x^r$. Since $x > 0$, any positive integer power of $x$ (i.e., $x^r$) will also be positive. The denominator $r!$ (factorial of $r$) is always positive for $r \ge 0$. Therefore, the sign of the entire term $T_{r+1}$ is determined solely by the sign of the binomial coefficient $\binom{n}{r}$.

3. Determine When the Binomial Coefficient $\binom{n}{r}$ Becomes Negative: For $\binom{n}{r}$ to be negative, the product of the terms in its numerator, $n(n-1)(n-2)\dots(n-r+1)$, must be negative. Our $n = \frac{27}{5} = 5.4$. This is a positive value. Let's examine the factors in the numerator:

• The first factor is $n = 5.4$, which is positive.
• Subsequent factors are $(n-1) = 4.4$, $(n-2) = 3.4$, and so on. These factors decrease as $r$ increases.
• Since $n$ is positive, the product will remain positive as long as all factors $(n-k)$ are positive.
• The product will become negative for the first time when one of these factors becomes negative. Since $n$ is positive and the factors are decreasing, the first factor to turn negative will be $(n-r+1)$.

Therefore, we need to find the smallest integer value of $r$ for which the factor $(n-r+1)$ becomes negative: $n - r + 1 < 0$ Substitute $n = \frac{27}{5}$: $\frac{27}{5} - r + 1 < 0$ To simplify, express $1$ as $\frac{5}{5}$: $\frac{27}{5} + \frac{5}{5} - r < 0$ $\frac{32}{5} - r < 0$ Convert the fraction to a decimal for easier comparison: $6.4 - r < 0$ Rearrange the inequality to solve for $r$: $6.4 < r$

4. Identify the Smallest Integer $r$ and the Corresponding Term Number: The inequality $r > 6.4$ tells us that $r$ must be an integer greater than $6.4$. The smallest integer value that satisfies this condition is $r=7$.

When $r=7$, the factor $(n-r+1)$ becomes $(5.4 - 7 + 1) = (5.4 - 6) = -0.6$, which is negative. This means that $\binom{27/5}{7}$ will be the first binomial coefficient in the expansion to have a negative value, because its numerator product will include one negative factor ($-0.6$) and all preceding factors are positive.

Since the term is $T_{r+1}$, for $r=7$, the term number is $T_{7+1} = T_8$.

Thus, the 8th term is the first negative term in the expansion of ${\left( {1 + x} \right)^{27/5}}$.

Relevant Tips and Common Mistakes to Avoid

• Understanding $n$: When $n$ is a positive integer, the binomial expansion is finite, and all terms are positive (if $x>0$). However, when $n$ is a fraction or a negative integer, the expansion is an infinite series, and terms can eventually become negative if $n$ is positive and fractional.
• Sign of $x$: Always check the sign of $x$. If $x$ were negative, $x^r$ would alternate in sign ($(-1)^r |x|^r$), which would affect the overall sign of the term, making the analysis more complex. Here, $x>0$ simplifies this, as $x^r$ is always positive.
• Term Index vs. Term Number: Remember that $r$ is the index in $\binom{n}{r} x^r$, but the actual term number is $r+1$. A common error is to directly state $r$ as the term number.

Summary

For the expansion of $(1+x)^{27/5}$ with $x>0$, the sign of a term $T_{r+1}$ is determined by its binomial coefficient $\binom{n}{r}$. Since $n = 5.4$, the coefficient $\binom{n}{r}$ becomes negative for the first time when the last factor in its numerator, $(n-r+1)$, becomes negative. Solving $n-r+1 < 0$ gives $r > 6.4$. The smallest integer $r$ satisfying this is $r=7$, which corresponds to the 8th term ($T_{7+1}$). Therefore, the 8th term is the first negative term.