Question

Find the value of ${}^n{C}_1{+}^{n}C5{+}^{n}C9{+}^n{C}_{13}.....$

• A. $2^{n-2}(1+sin\frac{n\pi }{4})$
• B. $2^{\frac{n-2}{2}}sin\frac{n\pi }{4}+2^{n-2}$
• C. $2^{\frac{n-1}{2}}sin\frac{n\pi }{4}+2^{n-2}$
• D. $2^{\frac{n}{2}}sin\frac{n\pi }{4}+2^{n-1}$

Answer 1

The correct option is B $2^{\frac{n-2}{2}}sin\frac{n\pi }{4}+2^{n-2}$

we have $(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2{+}^n{C}_3{x}^3{+}^n{C}_4{x}^4+....$
Putting x = - 1;
$2^n{=}^n{C}_0{+}^n{C}_1{+}^n{C}_2{+}^n{C}_3{+}^n{C}_4....---\left(1\right)$
Putting x = - 1;
$0{=}^n{C}_0{-}^n{C}_1{+}^n{C}_2{-}^n{C}_3{+}^n{C}_4-....---\left(2\right)$
subtracting (1)from (2); we get
$2^{n-1}{=}^n{C}_1{+}^{n}C3+n{C}_5{+}^n{C}_7{+}^n{C}_9+...----\left(3\right)$
Putting x = i;
$(1+i{)}^n{=}^n{C}_0{+}^n{C}_{1}i{+}^n{C}_2{i}^2{+}^n{C}_3{i}^3{+}^n{C}_4{i}^4+...\Rightarrow 2^{\frac{n}{2}}(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}{)}^n{=}^n{C}_0{+}^n{C}_{1}i{+}^n{C}_2(-1){+}^n{C}_3(-i){+}^n{C}_{4}i+...\Rightarrow 2^{\frac{n}{2}}(cos\frac{\pi }{4}+isin\frac{\pi }{4}{)}^n={(}^n{C}_0{-}^n{C}_2{+}^n{C}_4-...)+i{(}^n{C}_1{-}^n{C}_3{+}^{+}{C}_5-....)\Rightarrow 2^{\frac{n}{2}}{e}^{\frac{i\pi n}{4}}={(}^n{C}_0{-}^n{C}_2{+}^n{C}_4-...)+i{(}^n{C}_1{-}^n{C}_3{+}^{+}{C}_5-...)\Rightarrow 2^{\frac{n}{2}}(cos\frac{n\pi }{4}+isin\frac{n\pi }{4})={(}^n{C}_0{-}^n{C}_2{+}^n{C}_4-...)+i{(}^n{C}_1{-}^n{C}_3{+}^{+}{C}_5-....)$
Comparing the inaginary parts;
$\Rightarrow 2^{\frac{n}{2}}sin\frac{n\pi }{4}={(}^n{C}_1{-}^n{C}_3{+}^{+}{C}_5-....)----(4)$
Adding (3)and (4) ; we get
${}^n{C}_1{+}^n{C}_5{+}^n{C}_9{+}^n{C}_{13}....=2^{\frac{n-2}{2}}sin\frac{n\pi }{4}+2^{n-2}$