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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The sum of the coefficients of $x^r$ (where $r=0, 1, 2, \ldots, 15$) in the expansion of $(3x-1)^{15}$ is equal to the sum of the binomial coefficients of which of the following expansions?
$(a)\ (1+x)^{15}$
$(b)\ (1+x)^{16}+(1-x)^{16}$
$(c)\ (1+x)^{16}-(1-x)^{16}$

For $2 \le r \le n$,$\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$ is equal to

If $(1+x)^n = p_0 + p_1 x + p_2 x^2 + \ldots + p_n x^n$,then the value of $p_0 + p_3 + p_6 + \ldots$ is equal to:

If $^nC_r = C_r$ and $2 \frac{C_1}{C_0} + 4 \frac{C_2}{C_1} + 6 \frac{C_3}{C_2} + \dots + 2n \frac{C_n}{C_{n-1}} = 650$,then $^nC_2 =$

If $(1-x+x^2)^n = a_0 + a_1 x + \ldots + a_{2n} x^{2n}$,then the value of $a_0 + a_2 + a_4 + \ldots + a_{2n}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module