×

# Need help

Prove that

$\large 1 - ^{4n} C_1 - ^{4n} C_2 + ^{4n}C_3 + ^{4n}C_4 - ^{4n}C_5 - ^{4n}C_6 + \cdots + ^{4n}C_{4n} = (-4)^n .$

Note by Abdelfatah Teamah
8 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Notice that since $$\dbinom{4n}r=\dbinom {4n}{4n-r}$$, the terms $$\dbinom{4n}k$$ with $$k\equiv 1\pmod 4$$ and $$k\equiv -1\pmod 4$$ cancel out each other. The resulting series is,

$\binom{4n}0-\binom{4n}2+\dbinom{4n}4-\ldots-\binom{4n}{4n-2}+\binom{4n}{4n}=\sum_{k=0}^{2n}(-1)^k\binom{4n}{2k}$

Now notice that if $$\omega$$ be a primitive $$4^{\textrm{th}}$$ root of unity, say $$\omega=i$$, where $$i$$ is the imaginary unit (square root of -1), then the above series equals $$\frac 12\left((1+\omega)^{4n}+(1-\omega)^{4n}\right)$$.

With $$\omega=i$$, this equals $$\frac 12\left((1+i)^{4n}+(1-i)^{4n}\right)$$. Since $$(1+i)^4=(1-i)^4=-4$$, we have that the series equals $$\frac 12\left((-4)^n+(-4)^n\right)=(-4)^n$$, and we're done.

- 8 months ago

Thank you

- 7 months, 4 weeks ago

×