The number of terms in the expansion of $(x+y+z)^n$ ............ .
Given expansion is $(x+y+z)^n=[x+(y+z)]^n$.
$$\begin{aligned} {[x+(y+z)]^n={ }^n C_0 x^n+{ }^n C_1 x^{n-1}(} & (y) z) \\ & +{ }^n C_2 x^{n-2}(y+z)^2+\ldots+{ }^n C_n(y+z)^n \end{aligned}$$
$$\begin{aligned} \therefore \text { Number of terms } & =1+2+3+\ldots+n+(n+1) \\ & =\frac{(n+1)(n+2)}{2} \end{aligned}$$
In the expansion of $\left(x^2-\frac{1}{x^2}\right)^{16}$, the value of constant term is ............. .
Let constant be $T_{t+1}$.
$$\begin{aligned} T_{r+1} & ={ }^{16} C_r\left(x^2\right)^{16-r}\left(-\frac{1}{x^2}\right)^r \\ & ={ }^{16} C_r x^{32-2 r}(-1)^r x^{-2 r} \\ & ={ }^{16} C_r x^{32-4 r}(-1)^r \end{aligned}$$
For constant term, $\quad 32-4 r=0 \Rightarrow r=8$
$$\therefore \quad T_{8+1}={ }^{16} C_8$$
If the seventh term from the beginning and the end in the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$ are equal, then $n$ equals to ............ .
Given expansions is $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$.
$$T_7=T_{6+1}={ }^n C_6(\sqrt[3]{2})^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^6\quad \text{.... (i)}$$
Since, $T_7$ from end is same as the $T_7$ from beginning of $\left(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{2}\right)^n$.
Then, $$T_7={ }^n C_6\left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}(\sqrt[3]{2})^6\quad \text{.... (ii)}$$
Given that,
$${ }^n C_6(2)^{\frac{n-6}{3}}(3)^{-6 / 3}={ }^n C_6(3)^{-\frac{(n-6)}{3}} 2^{6 / 3}$$
$\Rightarrow\quad (2)^{\frac{n-12}{3}}=\left(\frac{1}{3^{1 / 3}}\right)^{n-12}$
which is true, when $\frac{n-12}{3}=0$.
$$\Rightarrow \quad n-12=0 \Rightarrow n=12$$
The coefficient of $a^{-6} b^4$ in the expansion of $\left(\frac{1}{a}-\frac{2 b}{3}\right)^{10}$ is $\qquad$
Given expansion is $\left(\frac{1}{a}-\frac{2 b}{3}\right)^{10}$.
Let $T_{t+1}$ has the coefficient of $a^{-6} b^4$.
$\therefore \quad T_{r+1}={ }^{10} C_r\left(\frac{1}{a}\right)^{10-r}\left(-\frac{2 b}{3}\right)^t$
For coefficient of $a^{-6} b^4, \quad 10-r=6 \Rightarrow r=4$
Coefficient of $a^{-6} b^4={ }^{10} C_4(-2 / 3)^4$
$$\therefore \quad =\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6!}{61 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{2^4}{3^4}=\frac{1120}{27}$$
Middle term in the expansion of $\left(a^3+b a\right)^{28}$ is ............ .
Given expansion is $\left(a^3+b a\right)^{28}$.
$$\begin{aligned} & \because \quad n=28 \quad \text{[even]}\\ & \therefore \quad \text { Middle term }=\left(\frac{28}{2}+1\right) \text { th term }=15 \text { th term } \\ & \therefore \quad T_{15}=T_{14+1} \\ & ={ }^{28} C_{14}\left(a^3\right)^{28-14}(b a)^{14} \\ & ={ }^{28} C_{14} a^{42} b^{14} a^{14} \\ & ={ }^{28} C_{14} a^{56} b^{14} \end{aligned}$$