Muszę udowodnić równość proszę o rozwiązanie dl n - l. naturalne:
\(\displaystyle{ \sum_{k=0}^{ [\frac{n}{2}]} {n\choose 2k} = 2^{n-1}}\)
Proszę o rozwiazania
Sigma-suma i silnia
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informacyjny01
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octahedron
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Sigma-suma i silnia
\(\displaystyle{ 2^n=(1+1)^n=\sum_{k=0}^{n} {n\choose k}1^k\cdot 1^{n-k} =\sum_{k=0}^{n} {n\choose k}\\
{n \choose k+1}= {n-1 \choose k+1}+{n-1 \choose k}\\
n=1:\\
\left[\frac{1}{2}\right]=0\\
\sum_{k=0}^{0} {1\choose 2k}={1\choose 0}=1=2^{1-1}\\
n>1:\\
n\text{ parzyste}:\\
\sum_{k=0}^{\left[\frac{n}{2}\right]} {n\choose 2k}={n\choose 0}+{n\choose 2}+{n\choose 4}+...+{n\choose n-2}+{n\choose n}=\\
={n-1\choose 0}+{n-1\choose 1}+{n-1\choose 2}+{n-1\choose 3}+{n-1\choose 4}+...+{n-1\choose n-3}+{n-1\choose n-2}+{n-1\choose n-1}=\sum_{k=0}^{n-1} {n-1\choose k}=2^{n-1}\\
n\text{ nieparzyste}:\\
\sum_{k=0}^{\left[\frac{n}{2}\right]} {n\choose 2k}={n\choose 0}+{n\choose 2}+{n\choose 4}+...+{n\choose n-3}+{n\choose n-1}=\\
={n-1\choose 0}+{n-1\choose 1}+{n-1\choose 2}+{n-1\choose 3}+{n-1\choose 4}+...+{n-1\choose n-4}+{n-1\choose n-3}+{n-1\choose n-2}+{n-1\choose n-1}=\sum_{k=0}^{n-1} {n-1\choose k}=2^{n-1}\\}\)
{n \choose k+1}= {n-1 \choose k+1}+{n-1 \choose k}\\
n=1:\\
\left[\frac{1}{2}\right]=0\\
\sum_{k=0}^{0} {1\choose 2k}={1\choose 0}=1=2^{1-1}\\
n>1:\\
n\text{ parzyste}:\\
\sum_{k=0}^{\left[\frac{n}{2}\right]} {n\choose 2k}={n\choose 0}+{n\choose 2}+{n\choose 4}+...+{n\choose n-2}+{n\choose n}=\\
={n-1\choose 0}+{n-1\choose 1}+{n-1\choose 2}+{n-1\choose 3}+{n-1\choose 4}+...+{n-1\choose n-3}+{n-1\choose n-2}+{n-1\choose n-1}=\sum_{k=0}^{n-1} {n-1\choose k}=2^{n-1}\\
n\text{ nieparzyste}:\\
\sum_{k=0}^{\left[\frac{n}{2}\right]} {n\choose 2k}={n\choose 0}+{n\choose 2}+{n\choose 4}+...+{n\choose n-3}+{n\choose n-1}=\\
={n-1\choose 0}+{n-1\choose 1}+{n-1\choose 2}+{n-1\choose 3}+{n-1\choose 4}+...+{n-1\choose n-4}+{n-1\choose n-3}+{n-1\choose n-2}+{n-1\choose n-1}=\sum_{k=0}^{n-1} {n-1\choose k}=2^{n-1}\\}\)