Question

The value of ${}^n{C}_0$ - ${}^n{C}_2$ + ${}^n{C}_4$ - ${}^n{C}_6$ . . . . .

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Consider the expansion of $(1+x{)}^n$.

$(1+x{)}^n$ = ${}^n{C}_0$ + ${}^n{C}_1$x + ${}^n{C}_2$ $x^2$ + ${}^n{C}_3$$x^3$ + ${}^n{C}_4$$x^4$...............

Put x = i,

⇒ $(1+i{)}^n$ = ${}^n{C}_0$ + ${}^n{C}_1$i - ${}^n{C}_2$ - ${}^n{C}_3$ i + ${}^n{C}_4$...................

(1 + i) = $\sqrt{2}$ $e^{\frac{i\pi }{4}}$

⇒$(1+i{)}^n$ = $2^{\frac{n}{2}}{e}^{i\frac{n\pi }{4}}$ = $2^{\frac{n}{2}}(icos\frac{n\pi }{4}+isin\frac{n\pi }{4})$

⇒$(1+i{)}^n$ = $2^{\frac{n}{2}}(cos\frac{n\pi }{4}+isin\frac{n\pi }{4})$

= n(${}^n{C}_0$ - ${}^n{C}_2$ + ${}^n{C}_4$...............) + i(${}^n{C}_1$ - ${}^n{C}_3$ + ${}^n{C}_5$...................)

Equating the real parts, we get

(${}^n{C}_0$ - ${}^n{C}_2$ + ${}^n{C}_4$.....................) = $2^{\frac{n}{2}}cos\frac{n\pi }{4}$