• # question_answer If ${{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+.....+{{C}_{n}}{{x}^{n}},$ then the value of ${{C}_{0}}-{{C}_{2}}+{{C}_{4}}-{{C}_{6}}+.....$is A) ${{2}^{n}}$ B) ${{2}^{n}}\cos \frac{n\pi }{2}$ C) ${{2}^{n}}\sin \frac{n\pi }{2}$ D) ${{2}^{n/2}}\cos \frac{n\pi }{4}$

Since ${{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+.....+{{C}_{n}}{{x}^{n}}$ Put$x=i$, on both the sides, we get ${{(1+i)}^{n}}=({{C}_{0}}-{{C}_{2}}+{{C}_{4}}-.....)+i({{C}_{1}}-{{C}_{3}}+{{C}_{5}}-.....)$                                                                          .....(i) Also, $1+i=\sqrt{2}\,\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)$ in amplitude modulus form   Þ${{(1+i)}^{n}}={{2}^{n/2}}{{\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)}^{n}}$ $={{2}^{n/2}}\left( \cos \frac{n\pi }{4}+i\sin \frac{n\pi }{4} \right)$                   ....(ii) Equating the real parts in (i) and (ii) we get, ${{C}_{0}}-{{C}_{2}}+{{C}_{4}}-{{C}_{6}}+.....={{2}^{n/2}}\cos \frac{n\pi }{4}$