Question

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

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

Answer 1

Answer: (A)

We know that,
$(1+x{)}^{3n}=1+{}^{3n}{C}_{1}x+{}^{3n}{C}_2{x}^2+\dots +{}^{3n}{C}_{3n}{x}^{3n}(1)$
$(1-x{)}^{3n}=1-{}^{3n}{C}_{1}x+{}^{3n}{C}_2{x}^2+\dots +(-1{)}^{3n}{}^{3n}{C}_{3n}{x}^{3n}(2)$
Subtracting (2) from (1) gives
$(1+x{)}^{3n}-(1-x{)}^{3n}=2\left[{}^{3n}{C}_{1}x+{}^{3n}{C}_3{x}^3+{}^{3n}{C}_5{x}^5+\dots \right]$
$=2x\left[{}^{3n}{C}_1+{}^{3n}{C}_3{x}^2+{}^{3n}{C}_5{x}^4+\dots \right]$
Putting $x=i\sqrt{3}$, we get
$(1+i\sqrt{3}{)}^{3n}-(1-i\sqrt{3}{)}^{3n}$
$=2i\sqrt{3}\left[{}^{3n}{C}_1-3\times {}^{3n}{C}_3+3^2\times {}^{3n}{C}_5\dots \right]$
Therefore, ${}^{3n}{C}_1-3^{3n}{C}_3+3^2\times {}^{3n}{C}_5\dots$
$=\frac{1}{2i\sqrt{3}}\left[(1+i\sqrt{3}{)}^{3n}-(1-i\sqrt{3}{)}^{3n}\right]$
$=\frac{1}{2i\sqrt{3}}\times 2^{3n}\left[\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)-{\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)}^{3n}\right]$
$=\frac{2^{3n-1}}{i\sqrt{3}}[(\cos n\pi +i\sin n\pi )-(\cos n\pi -i\sin n\pi )]$
$=\frac{2^{3n-1}}{i\sqrt{3}}2i\sin n\pi =0\text{ as }n$ is an integer.