Dowody.

[Artegor]

Udowodnij.

1.\(\sum_{k=0}^{n}(-1)^k { n\choose k }=0\)



2.\(\sum_{k=0}^{p} { n \choose k} { n-k \choose p-k } a^kb^{p-k}= {n \choose p}(a+b)^p\)

dla \(n \ge p \ge 0\)


3.\(\sum_{k=0}^{p} {n \choose k} {n-k \choose p-k} =2^p {n \choose p}\)

dla \(n \ge p \ge 0\)


4. \(\sum_{k=0}^{p} (-1)^k {n \choose k} {n-k \choose p-k}=0\)

dla \(n \ge p \ge 0\)

[eresh]

Udowodnij.

1.\(\sum_{k=0}^{n}(-1)^k { n\choose k }=0\)

korzystamy ze wzoru Newtona

\((a+b)^n=\sum_{k=0}^n{n\choose k}\cdot a^{n-k}\cdot b^k\)

jeśli \(a=1,b=-1\)
to:
\(\sum_{k=0}^n{n\choose k}\cdot 1^{n-k}\cdot (-1)^k=(1-1)^n\\
\sum_{k=0}^n{n\choose k}\cdot (-1)^k=0\)

[eresh]

Udowodnij.

2.\(\sum_{k=0}^{p} { n \choose k} { n-k \choose p-k } a^kb^{p-k}= {n \choose p}(a+b)^p\)

dla \(n \ge p \ge 0\)

\(\sum_{k=0}^{p} { n \choose k} { n-k \choose p-k } a^kb^{p-k}= \sum_{k=0}^p{n\choose p}\cdot {p\choose k}\cdot a^k\cdot b^{p-k}={n\choose p}\sum_{k=0}^p{p\choose k}\cdot a^k\cdot b^{p-k}={n\choose p}\cdot (a+b)^p\)

[eresh]

Udowodnij.
3.\(\sum_{k=0}^{p} {n \choose k} {n-k \choose p-k} =2^p {n \choose p}\)

dla \(n \ge p \ge 0\)

\(\sum_{k=0}^{p} {n \choose k} {n-k \choose p-k} =\sum_{k=0}^p{n\choose p}{p\choose k}={n\choose p}\sum_{k=0}^p{p\choose k}={n\choose p}\cdot (1+1)^p={n\choose p}\cdot 2^p\)

[eresh]

Udowodnij.
4. \(\sum_{k=0}^{p} (-1)^k {n \choose k} {n-k \choose p-k}=0\)

dla \(n \ge p \ge 0\)

\(\sum_{k=0}^{p} (-1)^k {n \choose k} {n-k \choose p-k}=\sum_{k=0}^p(-1)^k{n\choose p}{p\choose k}={n\choose p}\sum_{k=0}^p (-1)^k{p\choose k}={n\choose p}\cdot (1-1)^p=0\)