^{4n}C_0 + ^{4n}C_4 + ^{4n}C_8 + ... + ^{4n}C | vedclass

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(A) Let $S = ^{4n}C_0 + ^{4n}C_4 + ^{4n}C_8 + ... + ^{4n}C_{4n}$.
Consider the expansion of $(1+x)^{4n} = \sum_{r=0}^{4n} {^{4n}C_r} x^r$.
The sum o

Vedclass · Accepted Answer

$2^{4n - 2} + (-1)^n 2^{2n - 1}$

Vedclass · Answer

(A) Let $S = ^{4n}C_0 + ^{4n}C_4 + ^{4n}C_8 + ... + ^{4n}C_{4n}$.
Consider the expansion of $(1+x)^{4n} = \sum_{r=0}^{4n} {^{4n}C_r} x^r$.
The sum of terms with indices divisible by $4$ is given by the formula:
$S = \frac{1}{4} \left[ (1+1)^{4n} + (1-1)^{4n} + (1+i)^{4n} + (1-i)^{4n} \right]$.
$S = \frac{1}{4} \left[ 2^{4n} + 0 + (1+i)^{4n} + (1-i)^{4n} \right]$.
We know that $1+i = \sqrt{2} e^{i\pi/4}$ and $1-i = \sqrt{2} e^{-i\pi/4}$.
So,$(1+i)^{4n} = (\sqrt{2})^{4n} e^{in\pi} = 2^{2n} (\cos(n\pi) + i\sin(n\pi)) = 2^{2n} (-1)^n$.
Similarly,$(1-i)^{4n} = 2^{2n} (-1)^n$.
Thus,$(1+i)^{4n} + (1-i)^{4n} = 2 \cdot 2^{2n} (-1)^n = 2^{2n+1} (-1)^n$.
Substituting this back into the expression for $S$:
$S = \frac{1}{4} \left[ 2^{4n} + 2^{2n+1} (-1)^n \right] = 2^{4n-2} + 2^{2n-1} (-1)^n$.

$^{4n}C_0 + ^{4n}C_4 + ^{4n}C_8 + ... + ^{4n}C_{4n}$ is

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