If C_r = {}ⁿC_r , then C₀+C₄+C₈+C₁2} + ... =

Correct Answer is : 2n/2[cos⁡nπ4+2n/2−1]2{2ⁿ/2} ≤ft[ nπ/4 + 2ⁿ/2-1} ]}{2}22n/2[cos4nπ+2n/2−1]

{2ⁿ/2} ≤ft[ nπ/4 + 2ⁿ/2-1} ]}{2}

Correct Answer is : 2n/2[cos⁡nπ4+2n/2−1]2{2ⁿ/2} ≤ft[ nπ/4 + 2ⁿ/2-1} ]}{2}22n/2[cos4nπ+2n/2−1]

Test Index

AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Section: Mathematics

Question : 9 of 160

Marks: +1, -0

C_r

{}^{n}C_r

C_0+C_4+C_8+C_{12}

+ ... =

Solution:

From the given series,

(1)^{\frac{1}{4}}=(\cos 0+i\sin 0)^{\frac{1}{4}}

=(\cos 2r\pi+i\sin 2r\pi)^{\frac{1}{4}}

=\cos \frac{r\pi}{2}+i\sin \frac{r\pi}{2}

The four roots of unity for the following equation is

1, i, -1, -i=

1, \alpha, \alpha^{2}, \alpha^{3}

And

(1+x)^{n}=\sum\limits_{r=0}^{n} {}^{n}C_{r} x^{r}

For

x=1, (2)^{n}=\sum\limits_{r=0}^{n} {}^{n}C_{r}

For

x=\alpha(1+\alpha)^{n}=\sum\limits_{r=0}^{n} {}^{n}C_{r} \alpha^{r}

For

x=\alpha^{2}(1+\alpha^{2})^{n}=\sum\limits_{r=0}^{n} {}^{n}C_{r} \alpha^{2r}

For

x=\alpha^{3}(1+\alpha^{3})^{n}=\sum\limits_{r=0}^{n} {}^{n}C_{r} \alpha^{3r}

Add the above equations,

2^{n}+(1+\alpha)+(1+\alpha^{2})+(1+\alpha^{3})

=\sum\limits_{r=0}^{n} {}^{n}C_{r} (1+\alpha^{r}+\alpha^{2r}+\alpha^{3r})

Substitute the valeus in the above equation

4 ( {}^{n}C_{0}+ {}^{n}C_{4}+ {}^{n}C_{8}+\ldots\ldots )

=2^{n}+(1+i)^{n}+(1+i^{2})^{n}+(1+i^{3})^{n}

=2 \cdot 2^{\frac{n}{2}} \left[ \cos \frac{n\pi}{4}+2^{\frac{n}{2}-1} \right]

= \frac{2^{\frac{n}{2}} \left[ \cos \frac{n\pi}{4}+2^{\frac{n}{2}-1} \right]}{2}

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