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## Comments

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TopNewestJorok woy :v

obviously use binomial newton instead

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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

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