If $x^p$ occurs in the expansion of $\left(x^2+\frac{1}{x}\right)^{2 n}$, then prove that its coefficient is $\frac{2 n!}{\frac{(4 n-p)!}{3!} \frac{(2 n+p)!}{3!}}$.
Given expansion is $${\left( {{x^2} + {1 \over x}} \right)^{2n}}$$.
Let $x^p$ occur in the expansion of $${\left( {{x^2} + {1 \over x}} \right)^{2n}}$$.
$${T_{r + 1}} = {}^{2n}{C_r}{({x^2})^{2n - r}}{\left( {{1 \over x}} \right)^r}$$
$$ = {}^{2n}{C_r}{x^{4n - 2r}}{x^{ - r}} = {}^{2n}{C_r}{x^{4n - 3r}}$$
Let $$4n - 3r = p$$
$$ \Rightarrow 3r = 4n - p \Rightarrow r = {{4n - p} \over 3}$$
$\therefore$ Coefficient of $${x^p} = {}^{2n}{C_r} = {{(2n)!} \over {r!(2n - r)!}} = {{(2n)!} \over {\left( {{{4n - p} \over 3}} \right)!\left( {2n - {{4n - p} \over 3}} \right)!}}$$
$$ = {{(2n)!} \over {\left( {{{4n - p} \over 3}} \right)!\left( {{{6n - 4n + p} \over 3}} \right)!}} = {{(2n)!} \over {\left( {{{4n - p} \over 3}} \right)!\left( {{{2n + p} \over 3}} \right)!}}$$
Find the term independent of $x$ in the expansion of
$$\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$$
Given expansion is $\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$
Now, consider $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$
$$\begin{aligned} T_{r+1} & ={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r \\ & ={ }^9 C_r\left(\frac{3}{2}\right)^{9-r} x^{18-2 r}\left(-\frac{1}{3}\right)^r x^{-r}={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18-3 r} \end{aligned}$$
Hence, the general term in the expansion of $\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$
$$={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18-3 r}+{ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{19-3 r}+2 \cdot{ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{21-3 r}$$
For term independent of $x$, putting $18-3 r=0,19-3 r=0$ and $21-3 r=0$, we get
$$r=6, r=19 / 3, r=7$$
Since, the possible value of $r$ are 6 and 7.
Hence, second term is not independent of $x$.
$$\begin{aligned} & =\frac{9 \times 8 \times 7 \times 6!}{6!\times 3 \times 2} \cdot \frac{3^3}{2^3} \cdot \frac{1}{3^6}-2 \cdot \frac{9 \times 8 \times 7!}{7!\times 2 \times 1} \cdot \frac{3^2}{2^2} \cdot \frac{1}{3^7} \\ & =\frac{84}{8} \cdot \frac{1}{3^3}-\frac{36}{4} \cdot \frac{2}{3^5}=\frac{7}{18}-\frac{2}{27}=\frac{21-4}{54}=\frac{17}{54} \end{aligned}$$
18 The total number of terms in the expansion of $(x+a)^{100}+(x-a)^{100}$ after simplification is
If the integers $r>1, n>2$ and coefficients of $(3 r)$ th and $(r+2)$ nd terms in the Binomial expansion of $(1+x)^{2 n}$ are equal, then
The two successive terms in the expansion of $(1+x)^{24}$ whose coefficients are in the ratio $1:4$ are