Question

If $k=\frac{3n}{2},$ where $n$ is even positive integer, then $\sum _{r=1}^k{(-3)}^{r-1}{.}^{3n}{C}_{2r-1}=$

• A. $0$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is A $0$

We know that

${(1+x)}^{3n}=1{+}^{3n}{C}_1.x{+}^{3n}{C}_2.x^2+...{+}^{3n}{C}_{3n}.x^{3n}$ ...(1)

${(1-x)}^{3n}=1{-}^{3n}{C}_1.x{+}^{3n}{C}_2.x^2-...+{(-1)}^{3n}{C}_{3n}.x^{3n}$ ...(2)

Subtracting (2) from (1), we get

${(1+x)}^{3n}-{(1-x)}^{3n}=2[{}^{3n}{C}_1.x{+}^{3n}{C}_3.x^3{+}^{3n}{C}_5.x^5+...]=2[{}^{3n}{C}_1{+}^{3n}{C}_3.x{+}^{3n}{C}_5.x^4+...]$

Putting $x=i\sqrt{3}$, we get

${(1+i\sqrt{3})}^{3n}-{(1-i\sqrt{3})}^{3n}=2i\sqrt{3}[{}^{3n}{C}_1-{3.}^{3n}{C}_3+3^3{.}^{3n}{C}_5+....]$

${\therefore }^{3n}{C}_1-{3.}^{3n}{C}_3+3^3{.}^{3n}{C}_5+....=\frac{1}{2i\sqrt{3}}[{(1+i\sqrt{3})}^{3n}-{(1-i\sqrt{3})}^{3n}]$

$=\frac{1}{2i\sqrt{3}}.2^{3n}[{(\frac{1}{2}+\frac{i\sqrt{3}}{2})}^{3n}-{(\frac{1}{2}-\frac{i\sqrt{3}}{2})}^{3n}]$

$=\frac{2^{3n-1}}{i\sqrt{3}}[(\cos n\pi +i\sin n\pi )-(\cos n\pi -i\sin n\pi )]$

$=\frac{2^{3n-1}}{i\sqrt{3}}2i\sin n\pi =0$ as $n$ is an integer