The position of the term independent of $x$ in the expansion of $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}$ is ............. .
Given expansion is $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10}$.
Let the constant term be $T_{r+1}$.
Then,
$$\begin{aligned} T_{r+1} & ={ }^{10} C_r\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^2}\right)^r \\ & ={ }^{10} C_r \cdot x^{\frac{10-r}{2}} \cdot 3^{\frac{-10+r}{2}} \cdot 3^r \cdot 2^{-r} \cdot x^{-2 r} \\ & ={ }^{10} C_r x^{\frac{10-5 r}{2}} 3^{\frac{-10+3 r}{2}} 2^{-r} \end{aligned}$$
For constant term, $10-5 r=0 \Rightarrow r=2$
Hence, third term is independent of $x$./p>
If $25^{15}$ is divided by 13 , then the remainder is .............. .
Let
$$\begin{aligned} 25^{15} & =(26-1)^{15} \\ & ={ }^{15} C_0 26^{15}-{ }^{15} C_1 26^{14}+\ldots-{ }^{15} C_{15} \\ & ={ }^{15} C_0 26^{15}-{ }^{15} C_1 26^{14}+\ldots-1-13+13 \\ & ={ }^{15} C_0 26^{15}-{ }^{15} C_1 26^{14}+\ldots-13+12 \end{aligned}$$
It is clear that, when $25^{15}$ is divided by 13 , then remainder will be 12 .
The sum of the series $\sum_\limits{r=0}^{10}{ }^{20} C_r$ is $2^{19}+\frac{{ }^{20} C_{10}}{2}$.
The expression $7^9+9^7$ is divisible by 64 .
The number of terms in the expansion of $\left[\left(2 x+y^3\right)^4\right]^7$ is 8 .