Question

If${(1+x)}^n=\sum _{r=0}^n{C}_r{x}^r$, then the value of $C_0-C_2+C_4-C_6+C_8-C_{10}+...$ equals

• A. $2^{\frac{n}{2}}\cos \frac{nx}{4}$
• B. $C_1-C_3+C_5-C_7+....$
• C. $C_0+C_4+C_8+C_{12}+....$
• D. $2^{\frac{n}{2}}\sin \frac{nx}{4}$

Answer 1

Answer: (A)

$2^{\frac{n}{2}}\cos \frac{nx}{4}$

$(1+x{)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+...{}^n{C}_n{x}^n$
And
$(1-x{)}^n={}^n{C}_0-{}^n{C}_{1}x+{}^n{C}_2{x}^2+...(-1{)}^n{}^n{C}_n{x}^n$
Adding, we get
$(1+x{)}^n+(1-x{)}^n=2[{}^n{C}_0+{}^n{C}_2{x}^2+{}^n{C}_4{x}^4+{}^n{C}_6{x}^6+...]$
Substituting, $x=\sqrt{-1}=i$ we get
$(1+i{)}^n+(1-i{)}^n=2[{}^n{C}_0-{}^n{C}_2+{}^n{C}_4{x}^4-{}^n{C}_6{x}^6+...]$
$2^{\frac{n}{2}-1}[e^{\frac{in\pi }{4}}+e^{\frac{-in\pi }{4}}]={}^n{C}_0-{}^n{C}_2+{}^n{C}_4{x}^4-{}^n{C}_6{x}^6+...$
Hence
Solving the LHS, we get $2^{\frac{n}{2}}\cos (\frac{n\pi }{4})$