Binomial Theorem Test - 14

$$(1-x)\:^{25}$$ $$=1-\:^{25}C_{1}x+\:^{25}C_{2}x^2-\:^{25}C_{3}x^3+\:^{25}C_{4}x^4-\:^{25}C_{5}x^5...-\:^{25}C_{25}x^{25}$$
Putting $$x=1$$, we get
$$0=1-\:^{25}C_{1}+\:^{25}C_{2}-\:^{25}C_{3}+\:^{25}C_{4}-\:^{25}C_{5}...-\:^{25}C_{25}$$
Hence, sum of coefficients is $$0$$

$$T_r = {^{124}C_{r-1}}(\sqrt 3)^{125-r}(\sqrt[4]{5}) ^{r-1} $$
When both the terms are rational , $$T_r$$ will be rational. 
Hence, $$\dfrac{125-r}{2} $$ and $$\dfrac{r-1}{4} $$ both must be integers.
Therefore, $$r$$ must be of the form $$4k+1$$, where $$k$$ is an integer. 
There are $$125$$ terms in the expansion. Hence, $$k$$ can assume values from $$0$$ to $$31.$$ Hence, there are $$32$$ values of $$k$$ and $$32$$ rational terms in the expansion.
Hence, option $$B$$ is correct.

The real sol. is taking a binomial = $$(1+x)^{n} = ^{n}C_0 +  ^{n}C_1x + ^{n}C_2x^2 + ......$$
and now taking                              $$(1-x)^{n} = ^{n}C_0  - ^{n}C_1x + ^{n}C_2x^2 - ......$$ 
Now adding both and putting $$x = 1$$, we get $$2^n = 2( ^{n}C_0 + ^{n}C_2 + ^{n}C_4 + ....)$$
=> $$2^{n-1} $$ = required expression

We know $$T_{r+1}=\:^nC_{r}a^{n-r}b^r$$
Applying to the above question, we get
$$T_{2}=9x$$
Therefore $$9x=\:^nC_{1}ax$$
$$9=an$$ ...(i)
And $$T_{3}=27x^2$$
$$27x^2=\:^nC_{2}a^2x^2$$
$$\dfrac{n(n-1)}{2}a^2=27$$
$$n(n-1)a^2=54$$ ...(ii)
$$\therefore 9(an-a)=54$$ ...from (i)
$$an-a=6$$
$$9-6=a$$
$$\therefore a=3$$
Substituting in (i), we get
Therefore $$n=3$$ ...(from (i))

As in the hint , This is the first step. Now we have to find sum of $$r(n - r + 1)/r$$ = sum of $$(n - r + 1)$$ = 
$$n*n - n(n+1)/2 + n = n(n+1)/2$$. Here n= 15 which gives answer 120.

$$(1+x)^n=C_0+C_1 x+C_2 x^2+\dots +C_n x^n$$ ------(1)
$$\displaystyle(1+\frac{1}{x})^n=C_0+C_1\frac{1}{x}+C_2\frac{1}{x^2}+\dots +C_n\frac{1}{x^n}$$ -----(2)
Multiplying both sides, we get
$$\displaystyle\frac{(1+x)^{2n}}{(x)^n}=\sum{C_0^{2}}+x\sum{C_0C_1}+x^2\sum{C_0C_2}+ ....+x^r\sum{C_0C_r}+...$$
The various sums  are the coefficients of $$x^0,x,x^2,....x^r$$ in $$\displaystyle\frac{(1+x)^{2n}}{(x)^n}$$ or coefficients of $$x^n,x^{n+1},x^{n+2},....,x^{n+r}$$ in the expansion of $$(1+x)^{2n}$$ which occur in  $$T_{n+1},T_{n+2}....$$
Clearly the coefficent of $$x^n$$ in $$(1+x)^{2n}$$ is $$^{2n}C_n$$
$$\therefore  C_0^2+C_1^2+C_2^2 + \ldots\ldots +C_n^2 = ^{2n}C_n $$
Let $$S= C_0^2+3.C_1^2+5.C_2^2 + \ldots\ldots +(2n+1).C_n^2 $$ ----(3)
$$S=(2n+1).C_n^2+(2n-1).C_{n-1}^2+ \ldots\ldots+C_0^2$$ -----(4)
Adding (3) and (4) gives
$$2S=(2n+2)\left[C_0^2+C_1^2+C_2^2 + \ldots\ldots +C_n^2\right]$$
$$\therefore  S= (n+1). ^{2n}C_n $$
Hence, option C.

Substituting $$x=1$$ and $$y=1$$ in the above expression we get the sum of the binomial coefficients as
$$(5x-4y)^{n}$$
$$=(5-4)^{n}$$ (substituting x=1 and y=1)
$$=(1)^{n}$$
$$=1$$
Hence, answer is $$A.$$

To find the coefficient of $$x^{n}$$ in the expansion of $$\dfrac{(1+x)^{2n}}{1-x}$$
This can be written as $${(1+x)^{2n}}(1-x)^{-1}$$

$$= \left ( \binom{2n}{0} +  \binom{2n}{1}x +   \binom{2n}{2}x^{2} +   \binom{2n}{3}x^{3}+............+   \binom{2n}{n}  x^{n}  \right )  \left ( 1+ x+x^{2}+ x^{3}+x^{4}+.. . \right ) $$

On Multiplying this, we can find the coefficient of $$x^{n}$$ as summation  of
$$ = \binom{2n}{0} + \binom{2n}{1} + \binom{2n}{2} .......+\binom{2n}{n} $$
$$ =  3^{n} $$

$$ ^nC_r/^nC_{r-1} = n -r + 1/r $$. Now we have to find sum of r(n - r + 1)/r = sum of(n - r + 1) = 
n*n - n(n+1)/2 + n = n(n+1)/2. Here n= 15 which gives answer 120.

$$(1-x)^{10}$$ $$=1-\:^{10}C_{1}x+\:^{10}C_{2}x^2+...\:^{10}C_{10}x^{10}$$
Substituting $$x=1$$, we get sum of coefficients as
$$1-\:^{10}C_{1}+\:^{10}C_{2}+...\:^{10}C_{10}$$
$$=(1-1)^{10}$$
$$=0$$