Question

Find the value of ${}^n{C}_0$ +${}^n{C}_3$ +${}^n{C}_6$_{............................}

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

${}^n{C}_0$, ${}^n{C}_3$, ${}^n{C}_6$ are the coefficients in the expansion of $(1+x{)}^n$. We know how to find ${}^n{C}_0$ + ${}^n{C}_1$ + ${}^n{C}_2$...........

${}^n{C}_n$. We put x = 1 in the expansion of $(1+x{)}^n$. We know how to find ${}^n{C}_0$ + ${}^n{C}_2$ + ${}^n{C}_4$........... also we put x = -1

and x = 1 and add both. To get the term ${}^n{C}_0$, ${}^n{C}_3$, ${}^n{C}_6$..........we have to put x = 1 w, $w^2$ and add them.

This is because we know $w^{3n}$ = 1, and 1 + w + $w^2$ = 0. When we add the results we get after putting x = 1,

w and $w^2$, only the term ${}^n{C}_0$, ${}^n{C}_3$, ${}^n{C}_6$............ will remain.

Put x = w

⇒ $(1+w{)}^n$ = ${}^n{C}_0$ + ${}^n{C}_1$ w + ${}^n{C}_2$ $w^2$ + ${}^n{C}_3$ + ${}^n{C}_4$w + ${}^n{C}_5$ $w^2$ + ${}^n{C}_6$ .............(1)

Put x = $w^2$

⇒ $(1+w^2{)}^n$ = ${}^n{C}_0$ + ${}^n{C}_1$$w^2$ + ${}^n{C}_2$ $w^3$ + ${}^n{C}_3$ + ${}^n{C}_4$$w^2$ + ${}^n{C}_5$w + ${}^n{C}_6$.....................(2)

Put x = 1

⇒$(1+1{)}^n$ = ${}^n{C}_0$ + ${}^n{C}_1$ + ${}^n{C}_2$ + ${}^n{C}_3$ + ${}^n{C}_4$ + ${}^n{C}_5$ + ${}^n{C}_6$ ............(3)

(1) + (2) + (3)

⇒$(1+w{)}^n$ + $(1+w^2{)}^n$ + $2^n$

= 3${}^n{C}_0$ + ${}^n{C}_1$(1 + w + $w^2$) + ${}^n{C}_2$ (1 + w + $w^2$) + 3${}^n{C}_3$...................

= 3 (${}^n{C}_0$ + ${}^n{C}_3$ + ${}^n{C}_6$.................)

Let w = $\frac{1+i\sqrt{3}}{2}$ ⇒ $w^2$ = $\frac{-1-i\sqrt{3}}{2}$

⇒ 1 + w = $\frac{1+i\sqrt{3}}{2}$ = $e^{\frac{i\pi }{s}}$ and 1 + $w^2$ = $\frac{-1-i\sqrt{3}}{2}$ = $e^{\frac{-i\pi }{s}}$

⇒3(${}^n{C}_0$ + ${}^n{C}_3$ + ${}^n{C}_6$..................) = $e^{\frac{in\pi }{s}}$ + $e^{\frac{-in\pi }{s}}$ + $2^n$

⇒(${}^n{C}_0$ + ${}^n{C}_3$ + ${}^n{C}_6$.............) = $\frac{1}{3}$$(2cos\frac{n\pi }{3}+2^n)$