Question

$\left({}^m{C}_0{+}^m{C}_1{-}^m{C}_2{-}^m{C}_3\right)+\left({}^m{C}_4{+}^m{C}_5{-}^m{C}_6{-}^m{C}_7\right)+...=0$ if and only if for some positive integer $k,m=$

• A. $4k$
• B. $4k+1$
• C. $4k-1$
• D. $4k+2$

Answer 1

Answer: (C)

$4k-1$

Consider ${(\cos \theta -i\sin \theta )}^m{=}^m{C_0{\cos }^m\theta -}^m{C}_1{\cos }^{m-1}\theta i\sin \theta +...{+}^m{C}_m{(-i\sin \theta )}^m$ ....(1)

${(\cos \theta +i\sin \theta )}^m{=}^m{C_0{\cos }^m\theta +}^m{C}_1{\cos }^{m-1}\theta i\sin \theta +...{+}^m{C}_m{\left(i\sin \theta \right)}^m$ ....(2)

Adding (1) and (2), we get

$2\cos m\theta =2\left[{}^m{C}_0{\cos }^m\theta {-}^m{C}_2{\cos }^{m-2}\theta {\sin }^2\theta ...\right]$ ...(3)

Subtracting (3) and (4), we get

$2i\sin m\theta =2i\left[{}^m{C}_1{\cos }^{m-1}\theta i\sin \theta {-}^m{C}_3{\cos }^{m-3}\theta {\sin }^3\theta ...\right]$ ....(4)

Adding (3) and (4), we get

$\cos m\theta +\sin m\theta =\left[{}^m{C_0{\cos }^m\theta +}^m{C}_1{\cos }^{m-1}\theta i\sin \theta {-}^m{C_2{\cos }^{m-2}\theta {\sin }^2\theta -}^m{C}_3{\cos }^{m-3}\theta {\sin }^3\theta ...\right]$

$\Rightarrow \sqrt{2}\sin (m\theta +\frac{\pi }{4})=\left[{}^m{C_0{\cos }^m\theta +}^m{C}_1{\cos }^{m-1}\theta i\sin \theta {-}^m{C_2{\cos }^{m-2}\theta {\sin }^2\theta -}^m{C}_3{\cos }^{m-3}\theta {\sin }^3\theta ...\right]$

Putting $\theta =\frac{\pi }{4}$, we get

$\sqrt{2}\sin \left(\frac{(m+1)\pi }{4}\right)=\frac{1}{2^{\frac{m}{2}}}[\left({}^m{C}_0{+}^m{C}_1{-}^m{C}_2{-}^m{C}_3\right)+\left({}^m{C}_4{+}^m{C}_5{-}^m{C}_6{-}^m{C}_7\right)+...+\left({}^m{C}_{m-3}{+}^m{C}_{m-2}{-}^m{C}_{m-1}{-}^m{C}_m\right)]$

Hence $m+1=4k$, for given quantity to be $0$.

$\Rightarrow m=4k-1$ where $k\in N$