Given are the positive real numbers $a$ and $b$ and the natural number $n$.
Determine, in dependence on $a, b$, and $n$, the largest of the $n+1$ terms in the expansion of $(a+b)^{n}$.
The $k$-th term $\mathrm{G}(\mathrm{k})$ in the expansion of $(\mathrm{a}+\mathrm{b})^{\mathrm{n}}$ is given by the formula: $G(k)=\binom{n}{k-1} \cdot a^{n-k+1} \cdot b^{k-1}$, where $\binom{n}{0}=1$ and $1 \leq \mathrm{k} \leq \mathrm{n}+1$.
Since there are finitely many terms and every finite set of numbers (at least) contains a maximum element, it is investigated under what conditions the $k$-th element is maximal. For this, the following relationships must be satisfied simultaneously: $\binom{n}{k-1} \cdot a^{n-k+1} \cdot b^{k-1} \geq\binom{ n}{k} \cdot a^{n-k} \cdot b^{k}$ and $\binom{n}{k-1} \cdot a^{n-k+1} \cdot b^{k-1} \geq\binom{ n}{k-2} \cdot a^{n-k+2} \cdot b^{k-2}$, where in the latter case $k \geq 2$ must hold and one of these inequalities must be strict (simple proof!).
This leads on the one hand to $\frac{1}{n-k+1} \cdot a \geq \frac{1}{k} \cdot b$ and on the other hand to $\frac{1}{k-1} \cdot b \geq \frac{1}{n-k+2} \cdot a$, from which both $\mathrm{n}+1 \geq \mathrm{k} \geq \frac{n b+b}{a+b}$ and $1 \leq \mathrm{k} \leq \frac{n b+2 b+a}{a+b}=\frac{n b+b}{a+b}+1$ follow.
If $\mathrm{i}=\frac{n b+b}{a+b}$ is not an integer, then $\mathrm{G}(\mathrm{i})$ is the largest term, otherwise $\mathrm{G}(\mathrm{i})$ and $\mathrm{G}(\mathrm{i}+1)$ are maximal.
There cannot be any further maximal terms.