Obliczyć wartość wyrażenia

[zuza1414]

Będę bardzo wdzięczna za wyjaśnienie w jaki sposób obliczać wyrażenia takie jak te:

\(\displaystyle{ {101\choose 0}+ {101\choose 1}+...+{101\choose 50}}\)

[kerajs]

\(\displaystyle{ {101\choose 0}+ {101\choose 1}+...+{101\choose 50}=
\frac{1}{2} \left({101\choose 0}+ {101\choose 1}+...+{101\choose 50}+{101\choose 0}+ {101\choose 1}+...+{101\choose 50} \right) =\\= \frac{1}{2} \left({101\choose 0}+ {101\choose 1}+...+{101\choose 50}+{101\choose 101}+ {101\choose 100}+...+{101\choose 51} \right)= \frac{1}{2}\left( 1+1\right)^{101}


}\)

[zuza1414]

\(\displaystyle{ {101\choose 0}+ {101\choose 1}+...+{101\choose 50}=
\frac{1}{2} \left({101\choose 0}+ {101\choose 1}+...+{101\choose 50}+{101\choose 0}+ {101\choose 1}+...+{101\choose 50} \right) =\\= \frac{1}{2} \left({101\choose 0}+ {101\choose 1}+...+{101\choose 50}+{101\choose 101}+ {101\choose 100}+...+{101\choose 51} \right)= \frac{1}{2}\left( 1+1\right)^{101}


}\)

Dziękuje mam tylko pytanie dlaczego \(\displaystyle{ \frac{1}{2} }\)?

[Jan Kraszewski]

Bo \(\displaystyle{ x=\frac12(x+x).}\)

JK