help me

Note by Juang Bhakti Hastyadi
3 years, 4 months ago

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  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. 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 1

paragraph 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} \( \boxed{123} \)

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Jorok woy :v

obviously use binomial newton instead

Rimba Erlangga - 3 years, 4 months ago

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We consider the expansion \((1+x)^{4022}\) at \(x=1\) \[2^{4022} = \sum_{k=0}^{4022} {4022 \choose k}\] \[2^{4022} = {4022 \choose 0} + {4022 \choose 1} + {4022 \choose 2} + \ldots +{4022 \choose 4021} + {4022 \choose 4022}\]

Now, we make use of the following identity \({n \choose n-r} = {n \choose r}\), which gives

\[\begin{eqnarray} 2^{4022} & = & {4022 \choose 0} + {4022 \choose 1} + {4022 \choose 2} + \ldots + {4022 \choose 2009} + {4022 \choose 2010} + {4022 \choose 2011} \nonumber \\ \\ & & \quad + {4022 \choose 2012} + {4022 \choose 2013} + {4022 \choose 2014} + \ldots + {4022 \choose 4020} + {4022 \choose 4021} + {4022 \choose 4022} \nonumber \\ \\ & = & {4022 \choose 0} + {4022 \choose 1} + {4022 \choose 2} + \ldots + {4022 \choose 2009} + {4022 \choose 2010} + {4022 \choose 2011} \nonumber \\ \\ & & \quad + {4022 \choose 2010} + {4022 \choose 2009} + {4022 \choose 2008} + \ldots + {4022 \choose 2} + {4022 \choose 1} + {4022 \choose 0} \nonumber \\ \\ & = & 2\left({4022 \choose 0} + {4022 \choose 1} + {4022 \choose 2} + \ldots + {4022 \choose 2009} + {4022 \choose 2010}\right) + {4022 \choose 2011} \nonumber \\ \\ & = & 2\left(\sum_{k=0}^{2011} {4022 \choose k}\right) - {4022 \choose 2011} \nonumber \\ \end{eqnarray}\]

Thus, \[\sum_{k=0}^{2011} {4022 \choose k} = 2^{4021} + \frac{{4022 \choose 2011}}{2}\]

For further references, read - Investigating Binomial Theorem

Kishlaya Jaiswal - 3 years, 4 months ago

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