Middle term in the expansion of $\left(a^3+b a\right)^{28}$ is ............ .
Given expansion is $\left(a^3+b a\right)^{28}$.
$$\begin{aligned} & \because \quad n=28 \quad \text{[even]}\\ & \therefore \quad \text { Middle term }=\left(\frac{28}{2}+1\right) \text { th term }=15 \text { th term } \\ & \therefore \quad T_{15}=T_{14+1} \\ & ={ }^{28} C_{14}\left(a^3\right)^{28-14}(b a)^{14} \\ & ={ }^{28} C_{14} a^{42} b^{14} a^{14} \\ & ={ }^{28} C_{14} a^{56} b^{14} \end{aligned}$$
The ratio of the coefficients of $x^p$ and $x^q$ in the expansion of $(1+x)^{p+q}$ is .............. .
Given expansion is $(1+x)^{p+q}$.
$\therefore \quad$ Coefficient of $x^p={ }^{p+q} C_P$
and coefficient of $x^q={ }^{p+q} C_q$
$\therefore \quad \frac{{ }^{p+q} C_p}{{ }^{p+q} C_q}=\frac{p+q C_p}{p+q} C_p=1: 1$
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 $2^{19}+\frac{{ }^{20} C_{10}}{2}$.