Binomial Theorem Test - 25

$$(1+x)^{n}=a_{0}+a_{1}x+a_{2}x^2+...a_{n}x^{n}$$
For $$x=1$$ we get
$$2^{n}=a_{0}+a_{1}+a_{2}+...a_{n}$$ ...(i)
Substituting x=-1, we get
$$0=a_{0}-a_{1}+a_{2}+...-a_{n-1}+a_{n}$$ ...(ii)
Adding i and ii, we get
$$2^{n}=2(a_{0}+a_{2}+a_{4}+...a_{2r})$$
$$2^{n-1}=a_{0}+a_{2}+a_{4}+...a_{2r}$$
Now we know that $$a_{0}=\:^{n}C_{0}=1$$
By substituting, we get
$$a_{2}+a_{4}+...a_{2r}+..=2^{n-1}-1$$

Hence the required answer is $$2^{n-1}-1$$

Let
$$S_{n}=\frac{1}{2}\:^{10}C_{0}-\:^{10}C_{1}+...\:^{10}C_{10}2^{9}$$
$$2S_{n}=\:^{10}C_{0}-2\:^{10}C_{1}+2^{2}\:^{10}C_{2}+...\:^{10}C_{10}2^{10}$$
$$2S_{n}=(1-2)^{10}$$
$$2S_{n}=(1)$$
$$S_{n}=\frac{1}{2}$$

$$(1+2\sqrt{x})^{40}=1+\:^{40}C_{1}(2\sqrt{x})^{1}+\:^{40}C_{2}(2\sqrt{x})^{2}+\:^{40}C_{3}(2\sqrt{x})^{3}+...+\:^{40}C_{40}(2\sqrt{x})^{40}$$
Hence, all the odd terms will have integral powers of $$x$$.
Substituting $$x=1$$, we get
$$3^{40}=a_{0}+a_{1}+a_{2}+a_{3}...+a_{40}$$ ...(i)
Now consider,
$$(1-2\sqrt{x})^{40}=1-\:^{40}C_{1}(2\sqrt{x})^{1}+\:^{40}C_{2}(2\sqrt{x})^{2}-\:^{40}C_{3}(2\sqrt{x})^{3}...+\:^{40}C_{40}(2\sqrt{x})^{40}$$
Taking $$x$$ as $$1$$, we get
$$1=a_{0}-a_{1}+a_{2}-a_{3}...+a_{40}$$ ...(ii)
Adding i and, ii we get
$$3^{40}+1=2[a_{0}+a_{2}+a_{4}+a_{6}...+a_{40}]$$
$$\dfrac{3^{40}+1}{2}=a_{0}+a_{2}+a_{4}+a_{6}...+a_{40}$$
Hence the sum of the coefficients, of integral powers of $$x$$ will be
$$\dfrac{3^{40}+1}{2}$$
Hence, option D is correct

Using
$$\:^{n}C_{r}+\:^{n+1}C_{r}=\:^{n+1}C_{r+1}$$
$$\:^{10}C_{3}+\:^{11}C_{3}+\:^{12}C_{3}+\:^{13}C_{3}+...\:^{20}C_{3}$$
$$=[\:^{10}C_{4}+:^{10}C_{3}+\:^{11}C_{3}+\:^{12}C_{3}+\:^{13}C_{3}+...\:^{20}C_{3}]-\:^{10}C_{4}$$
$$=[\:^{11}C_{4}+:^{11}C_{3}+\:^{12}C_{3}+...\:^{20}C_{3}]-\:^{10}C_{4}$$
$$=[\:^{12}C_{4}+:^{12}C_{3}+\:^{13}C_{3}+...\:^{20}C_{3}]-\:^{10}C_{4}$$
:
:
:
$$=\:^{20}C_{4}+\:^{20}C_{3}-\:^{10}C_{4}$$
$$=\:^{21}C_{4}-\:^{10}C_{4}$$
$$=\:^{21}C_{17}-\:^{10}C_{6}$$

The last ten coefficients are 

We have, $$\sum _{ r=0 }^{ 10 }{ ^{ 20 }{ { C }_{ r } } } =^{ 20 }{ { C }_{ 0 } }+^{ 20 }{ { C }_{ 1 } }+...+^{ 20 }{ { C }_{ 10 } }$$

$$\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+\:^{20}C_{10}$$
$$=1+20+\dfrac{20(19)}{2}+\dfrac{20!}{10!(10!)}$$
$$=21+190+\dfrac{20!}{(10!)^{2}}$$
$$=211+\dfrac{20!}{(10!)^{2}}$$
Hence the answer is none of these.

In the expansion of $${ \left( 1+3\sqrt { 2 } x \right)  }^{ 9 }+{ \left( 1-3\sqrt { 2 } x \right)  }^{ 9 }$$

Simplifying, we get
$$(\:^{n+1}C_{1}+\:^{n+1}C_{2}+...\:^{n+1}C_{n})-(\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n})$$
$$=(\:^{n+1}C_{1}+\:^{n+1}C_{2}+...\:^{n+1}C_{n}+\:^{n+1}C_{n+1}-\:^{n+1}C_{n+1})-2^{n}$$
$$=2^{n+1}-2^{n}-\:^{n+1}C_{n+1}$$
$$=2^{n}(2-1)-1$$
$$=2^{n}-1$$

Given $$\displaystyle \left ( 1+x \right )^{n}=C_{0}+C_{1}x+C_{2}x^{2}+\cdots +C_{n}x^{n}$$ 
$$\displaystyle \therefore x^{2}\left ( 1+x \right )^{n}=x^{2}C_{0}+C_{1}x^{3}+C_{2}x^{4}+\cdots +C_{n}x^{n+2}\left ( * \right )$$ 
Now substituting $$x=1$$,$$\displaystyle \omega , \omega ^{2}in\left ( * \right ),$$ we get $$\displaystyle 2^{n}=C_{0}+C_{1}+C_{2}+C_{3}+\cdots +C_{n}\cdots \left ( 1 \right )$$
 $$\displaystyle \omega ^{2}\left ( 1+\omega  \right )^{n}=C_{0}\omega ^{2}+C_{1}\omega ^{3}+C_{2}\omega ^{4}+\cdots +C_{n}.\omega ^{n+2}$$...(2) 
$$\displaystyle \omega ^{4}\left ( 1+\omega ^{2} \right )^{n}=C_{0}\omega ^{4}+C_{1}\omega ^{6}+C_{2}\omega ^{8}+\cdots +C_{n}.\omega ^{2n+4}$$...(3) 
Now adding (1), (2) and (3) we get $$\displaystyle 2^{n}+\omega, ^{2}\left ( 1+\omega  \right )^{n}+\omega ^{4}\left ( 1+\omega ^{2} \right )^{n}$$ 
$$\displaystyle =C_{0}\left ( 1+\omega +\omega ^{2} \right )+C_{1}\left ( 1+\omega ^{3}+\omega ^{6} \right )+C_{2}\left ( 1+\omega +\omega ^{2} \right )$$
 $$\displaystyle +....=3\left ( C_{1}+C_{4}+C_{7}+\cdots  \right )$$ (other terms vanished).$$\displaystyle 3\left ( C_{1}+C_{4}+C_{7}+\cdots  \right )$$ $$\displaystyle =2^{n}+\omega ^{2}\left ( \cos\frac{n\pi }{2}+i \sin\frac{n\pi }{3} \right )+\omega ^{4}\left ( \cos\frac{n\pi }{3}-i \sin \frac{n\pi }{3} \right )$$ $$\displaystyle =2^{n}+\left ( \omega ^{2}+\omega  \right )\cos\frac{n\pi }{3}+i\left ( \omega ^{2}-\omega  \right )\sin \frac{n\pi }{3}$$ $$\displaystyle =2^{n}-\cos\frac{n\pi }{3}+i\left ( -\sqrt{3i} \right )\sin\frac{n\pi }{3}$$ As $$\displaystyle \omega =\frac{-1+i\sqrt{3}}{2}and \omega ^{2}=\frac{-1-i\sqrt{3}}{2}$$ $$\displaystyle \omega ^{2}-\omega =-\frac{1}{2}-\frac{i\sqrt{3}}{2}+\frac{1}{2}-\frac{i\sqrt{3}}{2}=i\sqrt{3}$$ $$\displaystyle \omega ^{2}-\omega =-\sqrt{3}$$ $$\displaystyle \therefore 3\left ( C_{1}+C_{4}+C_{7}\cdots  \right )=2^{n}-\cos\frac{n\pi }{3}+\sqrt{3}\sin \frac{n\pi }{3}$$ $$\displaystyle C_{1}+C_{4}+C_{7}+\cdots =\frac{1}{3}\left ( 2^{n}-\cos\frac{n\pi }{3}+\sqrt{3}\sin \frac{n\pi }{3} \right )$$