## Task B-4.4.

In the expansion of $(a+b)^{n}$, the third term is equal to $\frac{56}{9}$, the fourth term is $\frac{70}{3}$, and the binomial coefficients of the third and sixth terms are equal. Determine the numbers $a, b$, and $n$.
## Solution.

If the binomial coefficients of the third and sixth terms are equal, that is, if $\binom{n}{2}=\binom{n}{5}$, then due to the symmetry property of binomial coefficients, $n=7$.

The general term in the binomial expansion is $\binom{n}{k} a^{n-k} b^{k}$.

Therefore, the third term is $\binom{7}{2} a^{5} b^{2}=\frac{56}{9}$, and from this, $a^{5} b^{2}=\frac{8}{27}$.

The fourth term is $\binom{7}{3} a^{4} b^{3}=\frac{70}{3}$, and from this, $a^{4} b^{3}=\frac{2}{3}$.

If we divide the obtained equations, we get $\frac{a}{b}=\frac{a^{5} b^{2}}{a^{4} b^{3}}=\frac{8}{27} \cdot \frac{3}{2}=\frac{4}{9}$, so $a=\frac{4}{9} b . \quad 1$ point

From the first equation, it follows that $\left(\frac{4}{9} b\right)^{5} b^{2}=\frac{8}{27}$, and thus $b^{7}=\left(\frac{2}{3}\right)^{-7}$, and finally $b=\frac{3}{2}$.

1 point

$a=\frac{4}{9} b=\frac{4}{9} \cdot \frac{3}{2}=\frac{2}{3}$.

Note: A student who swaps $a$ and $b$ and solves the problem in this way gets all the points.