Binomial Theorem Test - 15

To find the coefficient of $$x^{n}$$ in the expansion of $$\dfrac{(1+x)^{n}}{1-x}$$.

Just calculate it by substituting the values of $$ ^5C_1 , ^5C_2$$  etc as $$5, 10$$ and just add them to get $$243$$.

Consider the expansion

$$(a-1)C_{1}-(a-2)C_{2}+(a-3)C_{3}-...(-1)^{n}(a-n)C_{n}$$
$$=[aC_{1}-aC_{2}+aC_{3}...]-[C_{1}-2C_{2}+C_{3}...]$$
$$=a(C_{1}-C_{2}+C_{3}...)-(C_{1}-2C_{2}+C_{3}...)$$
$$=a(1+(-1+C_{1}-C_{2}+C_{3}...))-(C_{1}-2C_{2}+C_{3}...)$$
$$=a+a(-1+C_{1}-C_{2}+C_{3}...)-(C_{1}-2C_{2}+C_{3}...)$$
$$=a+-a(1-x)^{n}|_{x=1}-n(1-x)^{n-1}|_{x=1}$$
$$=a$$

For $$(1+x)\:^{2n}$$
Middle term $$=\dfrac{N}{2}+1$$  since 2n is even
$$=\dfrac{2n}{2}+1$$
$$=(n+1)^{th}$$ term
Now consider the following
$$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$...(i)
Where $$T$$ represents the term
Here the middle term is the $$(n+1)^{th}$$ term
Hence $$r+1=n+1$$
$$r=n$$
Substituting in (i), we get
$$\:^{2n}C_{n}1^{2n-n}x^{n}$$
$$=\:^{2n}C_{n}x^{n}$$
Therefore $$a=\:^{2n}C_{n}$$ because a is the coefficient ...(a)

Consider $$(1+x)\:^{2n-1}$$
Since $$2n-1$$ is odd the middle terms are $$(\dfrac{N+1}{2})$$th and $$(\dfrac{N+1}{2}+1)$$ th terms
Now consider the following
$$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$...(ii)
Where $$T$$ represents the term
Here the middle terms are the $$(n)^{th}$$ term and $$(n+1)^{th}$$ term.
Hence $$r+1=n+1$$ and $$r+1=n$$
$$r=n$$ and $$r=n-1$$
For $$r=n$$
$$T_{r+1}=\:^{2n-1}C_{n}1^{n-1}x^{n}$$
$$=\:^{2n-1}C_{n}x^{n}$$
Hence, $$b=\:^{2n-1}C_{n}$$, since $$b$$ is the coefficient ...(b)

Similarly for $$r=n-1$$
$$T_{r+1}=\:^{2n-1}C_{n-1}1^{n}x^{n-1}$$
$$=\:^{2n-1}C_{n-1}x^{n-1}$$
Hence $$c=\:^{2n-1}C_{n-1}$$ since $$c$$ is the coefficient ...(c)
From $$a,b,c$$, we get
$$\:^{2n}C_{n}=\:^{2n-1}C_{n}+\:^{2n-1}C_{n-1}$$
$$a=b+c$$

$$(1+2^{1/2}+3^{1/3})^{6}$$
$$=[(1+2^{1/2}+3^{1/3})^{3}]^{2}$$
$$=[(1+(2^{1/2}+3^{1/3})^{3}+3(2^{1/2}+3^{1/3})^{2}+3(2^{1/2}+3^{1/3})]^{2}$$
$$=1+(2^{1/2}+3^{1/3})^{6}+9(2^{1/2}+3^{1/3})^{4}+9(2^{1/2}+3^{1/3})^{2}+2(2^{1/2}+3^{1/3})^{3}+\\ 6(2^{1/2}+3^{1/3})^{5}+18(2^{1/2}+3^{1/3})^{3}+6(2^{1/2}+3^{1/3})$$
Hence the rational terms are
$$1,24,4,12,6,9,8$$
Hence, there are total $$7$$ rational terms

$$(1-2x^3+3x^5)(1+x^{-1})^{8}$$

Given, $$(1+x)\:^n(1+\dfrac{1}{x})\:^n$$
$$=\dfrac{(1+x)^{2n}}{x^{n}}$$ ...(i)
Therefore$$T_{r+1}=\dfrac{\:^{2n}C_{r}x^{r}}{x^{n}}$$
$$=\:^{2n}C_{r}x^{r-n}$$ ...(ii)
For coefficient of $$ x^{-1}$$
$$r-n=-1$$
$$r=n-1$$
Substituting in (ii), we get
$$T_{n-2}=\:^{2n}C_{n-1}x^{-1}$$
$$=\dfrac{2n!}{(n+1)!(n-1)!\:x}$$

Given, $$(1-3x+3x^2-x^3)\:^{2n}$$

$$(x^2+x^{-1})^{2n}$$
$$T_{r+1}=\:^{2n}C_{r}x^{4n-3r}$$ ...(i)
For the coefficient of $$x^p$$
$$4n-3r=p$$
$$\dfrac{4n-p}{3}=r$$
Substituting in equation (i), we get
$$T_{(\frac{4n-p}{3})+1}=\:^{2n}C_{\frac{4n-p}{3}}x^{p}$$
Hence, coefficient of $$x^p$$ in the above binomial expansion is $$\:^{2n}C_{\frac{4n-p}{3}}$$.