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How should I prove the maximum term in the expansion of $(x+a)^n$ where $ax>0$ is the term $C(n,r)x^{(n-r)}a^r$ for which $r= \left[\cfrac{(n+1)}{(n/a)+1} \right]$ ?

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2 Answers 2

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Hint: you want to maximize $$ u_r = \binom nr x^{n-r}a^r $$Every term is positive, and $$ \frac{u_{r+1}}{u_r} = \frac{a/x}{(r+1)/(n-r)} $$

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Consider two consecutive terms in the expansion of $(x + a)^n$: $T_{r+1}$ and $T_r$

Now, take the ratio of these two terms, you get:

$$\begin{align} & \left| \cfrac{T_{r+1}}{T_r} \right| = \left| \cfrac{ \binom{n}{r} (a^{n-r} x^r) }{\binom{n}{r-1} a^{n - r + 1} x^{r-1}} \right| = \left| \cfrac{\binom{n}{r}}{\binom{n}{r-1}} \right| \left|\cfrac{x}{a} \right| \\ = & \left| \cfrac{n - r + 1}{r} \right| \left| \cfrac{x}{a} \right| = \color{blue}{\left| \cfrac{n+1}{r} - 1 \right| \left| \cfrac{x}{a} \right| }\end{align} $$

Thus, $$\begin{align} & |T_{r+1}| > |T_r| \ \text{if} \ \color{blue}{\left\{\cfrac{n+1}{r} - 1 \right\} \left| \cfrac{x}{a} \right|} > 1 \ \\ & \text{that is: } \ \cfrac{n+1}{r} > 1 + \left| \cfrac{a}{x} \right| \implies \cfrac{n+1}{1 + \left|\cfrac{a}{x} \right| } > r \tag{1} \end{align} $$

**Since, $n>r$ thus, $\cfrac{n+1}{r} - 1$ must be positive. Thus, $T_{r+1}$ will be the greatest term, if $r$ has the greatest value as per the equation (1) **

EDIT :

From considering $T_{r+1} \ge T_r $ , we have: $$\begin{align} & \cfrac{T_{r+1}}{T_r} \ge 1 \\ & \cfrac{n-r+1}{r} \times x \ge 1 \implies \cfrac{r}{n-r+1} \le x \tag{2} \end{align}$$

Now, consider this: $T_{r+1} \ge T_{r+2} $, we have: $$\begin{align} & \cfrac{T_{r+1}}{T_{r+2}} \ge 1 \\ & \cfrac{(r+1)}{(n-r)x} \ge 1 \\ & x \le \cfrac{(r+1)}{n-r} \end{align} \tag{3} $$

Now, from equations $(2) \ \& \ (3) $ we get: $$\cfrac{r}{n-r+1} \le x \le \cfrac{r+1}{n-r} \tag{4} $$

(For equations 2 and 3, if x is negative then we gave to take its absolute value.)

Equation 4 is applicable for x to be non-positive integer too. It is valid for negative integer or fraction also.

Finally, we conclude: $$\cfrac{r}{n - r + 1} \le |x| \le \cfrac{r+1}{n-r} \tag{Final}$$

Thus, for the greatest value of $|x|$, we have: $$|x|_{\text{Max.}} = \cfrac{r+1}{n-r} $$

Now, let us try to simplify the Final equation further: $$\cfrac{r}{n - r + 1} \le |x| \le \cfrac{r+1}{n-r} $$

Consider the first two expressions of the inequality first: $$\begin{align} & \cfrac{r}{n - r + 1} \le |x| \\ & r \le |x| (n- r + 1) \\ & r \le |x|(n) - |x|(r) + |x| \\ & r + |x|(r) \le |x|(n) + |x| \\ & r(1 + |x|) \le |x|(n+1) \\ & \color{blue}{r \le \cfrac{|x|(n+1)}{1 + |x|} } \tag{Final(1.1)}\end{align} $$

Now, consider the next part of the inequality: $$\begin{align} & |x| \le \cfrac{r+1}{n-r} \\ & |x|(n-r) \le r +1 \\ & |x| (n) - |x| (r) \le r + 1 \\ & |x|(n) - 1 \le r + |x| (r) \\ & |x|(n) - 1 \le r(1 + |x|) \\ & \color{blue}{r \ge \cfrac{n(|x|) - 1}{1 + |x|} } \tag{Final(1.2)}\end{align}$$

So, you have another inequality now: $$\color{red}{\cfrac{n|x|-1}{|x| + 1} \le r \le \cfrac{(n+1)|x|}{|x| + 1}} $$

Clearly, the maximum value of $r$ will be: $$\color{red}{\bf{r = \cfrac{(n+1)x}{x+1} } } $$

Remember that this is for the binomial expansion of $(1 + x)^n$ and not for $(a + x)^n$.

This all was just to add this to your knowledge.

COMING-BACK TO POINT :

Recall that, we found equation (1) which is: $$\cfrac{n + 1}{1 + \cfrac{|a|}{|x|} } > r$$

Implies, $$r < \cfrac{n+1}{|x| + |a| } \times |x| $$

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  • $\begingroup$ Dear Kushashwa, how should I prove the greatest value of the binomial coefficient $C(n,r)$ occurs for $r=\left[\cfrac{(n+1)}{2}\right]$ from above result? $\endgroup$ Jul 3, 2014 at 13:13

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