Explanation
Given $$\displaystyle \left ( 1+x \right )^{n}=C_{0}+C_{1}x+C_{2}x^{2}+\cdots +C_{n}x^{n}$$ $$\displaystyle \therefore x^{2}\left ( 1+x \right )^{n}=x^{2}C_{0}+C_{1}x^{3}+C_{2}x^{4}+\cdots +C_{n}x^{n+2}\left ( * \right )$$ Now substituting $$x=1$$,$$\displaystyle \omega , \omega ^{2}in\left ( * \right ),$$ we get $$\displaystyle 2^{n}=C_{0}+C_{1}+C_{2}+C_{3}+\cdots +C_{n}\cdots \left ( 1 \right )$$ $$\displaystyle \omega ^{2}\left ( 1+\omega \right )^{n}=C_{0}\omega ^{2}+C_{1}\omega ^{3}+C_{2}\omega ^{4}+\cdots +C_{n}.\omega ^{n+2}$$...(2) $$\displaystyle \omega ^{4}\left ( 1+\omega ^{2} \right )^{n}=C_{0}\omega ^{4}+C_{1}\omega ^{6}+C_{2}\omega ^{8}+\cdots +C_{n}.\omega ^{2n+4}$$...(3) Now adding (1), (2) and (3) we get $$\displaystyle 2^{n}+\omega, ^{2}\left ( 1+\omega \right )^{n}+\omega ^{4}\left ( 1+\omega ^{2} \right )^{n}$$ $$\displaystyle =C_{0}\left ( 1+\omega +\omega ^{2} \right )+C_{1}\left ( 1+\omega ^{3}+\omega ^{6} \right )+C_{2}\left ( 1+\omega +\omega ^{2} \right )$$ $$\displaystyle +....=3\left ( C_{1}+C_{4}+C_{7}+\cdots \right )$$ (other terms vanished).$$\displaystyle 3\left ( C_{1}+C_{4}+C_{7}+\cdots \right )$$ $$\displaystyle =2^{n}+\omega ^{2}\left ( \cos\frac{n\pi }{2}+i \sin\frac{n\pi }{3} \right )+\omega ^{4}\left ( \cos\frac{n\pi }{3}-i \sin \frac{n\pi }{3} \right )$$ $$\displaystyle =2^{n}+\left ( \omega ^{2}+\omega \right )\cos\frac{n\pi }{3}+i\left ( \omega ^{2}-\omega \right )\sin \frac{n\pi }{3}$$ $$\displaystyle =2^{n}-\cos\frac{n\pi }{3}+i\left ( -\sqrt{3i} \right )\sin\frac{n\pi }{3}$$ As $$\displaystyle \omega =\frac{-1+i\sqrt{3}}{2}and \omega ^{2}=\frac{-1-i\sqrt{3}}{2}$$ $$\displaystyle \omega ^{2}-\omega =-\frac{1}{2}-\frac{i\sqrt{3}}{2}+\frac{1}{2}-\frac{i\sqrt{3}}{2}=i\sqrt{3}$$ $$\displaystyle \omega ^{2}-\omega =-\sqrt{3}$$ $$\displaystyle \therefore 3\left ( C_{1}+C_{4}+C_{7}\cdots \right )=2^{n}-\cos\frac{n\pi }{3}+\sqrt{3}\sin \frac{n\pi }{3}$$ $$\displaystyle C_{1}+C_{4}+C_{7}+\cdots =\frac{1}{3}\left ( 2^{n}-\cos\frac{n\pi }{3}+\sqrt{3}\sin \frac{n\pi }{3} \right )$$
$$\:^{n}C_{r+1}+\:^{n}C_{r}$$ $$=\:^{n+1}C_{r+1}$$ Hence simplifying the terms, we get $$\:^{n+1}C_{1}.\:^{n+1}C_{2}....\:^{n+1}C_{n}$$ Now $$\:^{n+1}C_{1}$$ $$=\dfrac {(n+1)!}{n!.1!}$$ $$=\dfrac {(n+1)}{n}\:^{n}C_{1}$$ Similarly $$\:^{n+1}C_{2}$$ $$=\dfrac {(n+1)!}{(n-1)!.2!}$$ $$=\dfrac {(n+1)}{(n-1)}\:^{n}C_{2}$$ Hence substituting, in the above expression, we get $$\dfrac {(n+1)^{n}}{n!}(\:^{n}C_{0}.\:^{n}C_{1}...\:^{n}C_{n})$$ Comparing coefficients, we get $$K=\dfrac {(n+1)^{n}}{n!}$$
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