1. Tardigrade
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  4. If k=(3 n/2), where n is an even positive integer, then displaystyle∑r=1k(-3)r-1 ⋅ 3 n C 2 r-1=

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

Complex Numbers and Quadratic Equations

Solution:

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