Question

The value of ${2000}_{C_2}+{2000}_{C_5}+{2000}_{C_8}+...+{2000}_{C_{2000}}=?$

A. $\frac{2^{1999}+1}{3}$

B. $\frac{2^{1999}-1}{3}$

C. $\frac{2^{2000}+1}{3}$

D. $\frac{2^{2000}-1}{3}$

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

We know that.

$(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2{+}^n{C}_3{x}^3+....{.}^n{C}_n{x}^n$

Now, put $x=1$

$\Rightarrow (1+1{)}^n{=}^n{C}_0{+}^n{C}_1(1\left){+}^n{C}_2\right(1{)}^2{+}^n{C}_3\left(1\right)3+....{.}^n{C}_n(1{)}^n$

$\Rightarrow 2^n{=}^n{C}_0{+}^n{C}_1{+}^n{C}_2{+}^n{C}_3+....{.}^n{C}_n$-----(1)

Put $x=\omega$ in $(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2{+}^n{C}_3{x}^3+....{.}^n{C}_n{x}^n$

$\Rightarrow (1+\omega {)}^n{=}^n{C}_0{+}^n{C}_1(\omega \left){+}^n{C}_2\right(\omega {)}^2{+}^n{C}_3(\omega {)}^3+....{.}^n{C}_n(\omega {)}^n$

$\Rightarrow \omega (1+\omega {)}^n={\omega }^n{C}_0{+}^n{C}_1(\omega {)}^2{+}^n{C}_2\left(\omega {)}^3{+}^n{C}_3\right(\omega {)}^4+....{.}^n{C}_n(\omega {)}^{n+1}$-----(2)

Put $x={\omega }^2$ in $(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2{+}^n{C}_3{x}^3+....{.}^n{C}_n{x}^n$

$\Rightarrow (1+{\omega }^2{)}^n{=}^n{C}_0{+}^n{C}_1({\omega }^2\left){+}^n{C}_2\right(\omega {)}^4{+}^n{C}_3(\omega {)}^6+....{.}^n{C}_n(\omega {)}^{2n}$

$\Rightarrow {\omega }^2(1+{\omega }^2{)}^n{=}^n{C}_0{\omega }^2{+}^n{C}_1({\omega }^4\left){+}^n{C}_2\right(\omega {)}^6{+}^n{C}_3(\omega {)}^8+....{.}^n{C}_n(\omega {)}^{2n+2}$

-----(3)

Now add eq (1), eq(2) and eq(3),

$2^n+\omega (1+\omega {)}^n+{\omega }^2(1+\omega {)}^n{=}^n{C}_0(1+\omega +{\omega }^2){+}^n{C}_1(1+\omega +{\omega }^2)+3^n{C}_2{+}^n{C}_3(1+\omega +{\omega }^2){}^{}{+}^n{C}_4(1+\omega +{\omega }^2)+3^n{C}_2+.........+3^n{C}_{2+3k}$

Since, $1+\omega +{\omega }^2=0$

$\Rightarrow 2^n+\omega (1+\omega {)}^n+{\omega }^2(1+\omega {)}^n{=}^n{C}_0\left(0\right){+}^n{C}_1\left(0\right)+3^n{C}_2{+}^n{C}_3\left(0\right){+}^n{C}_4\left(0\right)+3^n{C}_2+.........+3^n{C}_{2+3k}$

$\Rightarrow 2^n+\omega (1+\omega {)}^n+{\omega }^2(1+\omega {)}^n=3^n{C}_2+3^n{C}_5+.........+3^n{C}_{(2+3k)}$

$\Rightarrow \frac{2^n+\omega (1+\omega {)}^n+{\omega }^2(1+\omega {)}^n}{3}{=}^n{C}_2{+}^n{C}_5+.........{+}^n{C}_{(2+3k)}$

Now, lets take $n=2000$

$\Rightarrow \frac{2^{2000}+\omega (1+\omega {)}^{2000}+{\omega }^2(1+{\omega }^2{)}^{2000}}{3}={}^{2000}{C}_2{+}^{2000}{C}_5+.........{+}^{2000}{C}_{2000}$

$\Rightarrow \frac{2^{2000}+\omega \left(\right(1+\omega {)}^2{)}^{1000}+{\omega }^2(-\omega {)}^{2000}}{3}={}^{2000}{C}_2{+}^{2000}{C}_5+.........{+}^{2000}{C}_{2000}$

$\Rightarrow \frac{2^{2000}+\omega (-{\omega }^2{)}^{2000}+{\omega }^2(\omega {)}^{2000}}{3}={}^{2000}{C}_2{+}^{2000}{C}_5+.........{+}^{2000}{C}_{2000}$

$\Rightarrow \frac{2^{2000}+\omega (\omega {)}^{4000}+{\omega }^2(\omega {)}^{2000}}{3}={}^{2000}{C}_2{+}^{2000}{C}_5+.........{+}^{2000}{C}_{2000}$

$\Rightarrow \frac{2^{2000}+\omega \left(\omega \right)+{\omega }^2(\omega {)}^2}{3}={}^{2000}{C}_2{+}^{2000}{C}_5+.........{+}^{2000}{C}_{2000}$

$\Rightarrow \frac{2^{2000}+{\omega }^2(1+{\omega }^2)}{3}={}^{2000}{C}_2{+}^{2000}{C}_5+.........{+}^{2000}{C}_{2000}$

$\Rightarrow \frac{2^{2000}+{\omega }^2(-\omega )}{3}={}^{2000}{C}_2{+}^{2000}{C}_5+.........{+}^{2000}{C}_{2000}$

$\Rightarrow \frac{2^{2000}-{\omega }^3}{3}={}^{2000}{C}_2{+}^{2000}{C}_5+.........{+}^{2000}{C}_{2000}$

$\Rightarrow \frac{2^{2000}-1}{3}={}^{2000}{C}_2{+}^{2000}{C}_5+.........{+}^{2000}{C}_{2000}$

Hence, option D is correct.