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Show that the coefficient of the middle term in the expansion of $(1+x)^{2n}$ is equal to the sum of the coefficients of the two middle terms in the expansion of $(1+x)^{2n-1}$.

If the eleventh term in the binomial expansion of $(x+a)^{15}$ is the geometric mean of the eighth and twelfth terms,then the greatest term in the expansion is

If the coefficient of the $3^{\text{rd}}$ term from the beginning in the expansion of $\left(ax^2 - \frac{8}{bx}\right)^9$ is equal to the coefficient of the $3^{\text{rd}}$ term from the end in the expansion of $\left(ax - \frac{2}{bx^2}\right)^9$,then the relation between $a$ and $b$ is:

The coefficient of $x^6$ in the binomial expansion of ${\left( \frac{4x^2}{3} - \frac{3}{2x} \right)^9}$ is

In the binomial expansion of $(a - b)^n, n \ge 5,$ the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $a/b$ equals:

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