Question

$If(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+.......C_n{x}^n....(i)$
then sum of series $C_0+C_k+C_{2k}+$....can be obtained by putting all roots of equation $x^k-1=0$ in (i) & then adding vertically:
for example: sum of $C_0+C_2+C_4$.....can be obtained by putting all roots of equation $x^2=1$
i.e. $x=\pm 1$ in (i)
At $x=1C_0+C_1+C_2..........C_n=2^n$
$x=-1C_0-C_1+C_2-C_3...........=0$
Adding we get $C_0+C_2+C_4.....=2^{n-1}$

Now answer the folloiwng

If n is a multiple of 3, then $C_0+C_3+C_6+................$ equals

• A. $\frac{2^n+2}{3}$
• B. $\frac{2^n-2}{3}$
• C. $\frac{2^n+2(-1{)}^n}{3}$
• D. $\frac{2^n-2(-1{)}^n}{3}$

Answer 1

The correct option is C

$\frac{2^n+2(-1{)}^n}{3}$

$x^3-1=0$ has roots$1,\omega ,{\omega }^2$
At $x=1C_0+C_1+C_2...........C_n=2^n......\left(i\right)x=\omega C_0+C_1\omega +C_2{\omega }^2......+C_n{\omega }^n=(1+\omega {)}^n......(ii)x={\omega }^2{C}_0+C_1{\omega }^2+C_2{\omega }^4......+C_n{\omega }^{2n}=(1+{\omega }^2{)}^n.....\left(iii\right)$

Adding (i), (ii) and (iii) , we get

$3(C_0+C_3+C_6.....)=2^n+(1+\omega {)}^n+(1+{\omega }^2{)}^n[1+\omega +{\omega }^2=0]C_0+C_3+C_6......=\frac{2^n+(-{\omega }^2{)}^n+(-\omega {)}^n}{3}$

If n is a multiple of 3

then $C_0+C_3.............=\frac{2^n+2(-1{)}^n}{3}$

Now $x^4-1=0$

has roots $x=\pm 1,\pm i$

$\therefore$ sum of values of x = 0