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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