1)Doprowadz do najprostszej postaci, podaj zalozenia:
\(\displaystyle{ \frac{(2n-1)! (n+2)!}{n! (2n+2)!}}\)
2)Oblicz:
\(\displaystyle{ {25\choose 23}}\) + \(\displaystyle{ {25\choose 24}}\)
3)Zapisz w postaci sumy korzystajac ze wzoru dwumianu newtona:
\(\displaystyle{ \left(x^{2}+\frac{1}{x^{2}}\right)^{5}}\); (x0)
silnia, dwumian newtona
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natkoza
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- Rejestracja: 11 kwie 2007, o 18:49
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silnia, dwumian newtona
2.
\(\displaystyle{ {25\choose 23} +{25\choose 24}=\frac{25!}{23!\cdot 2!}+\frac{25!}{24!}=\frac{23!\cdot 24\cdot 25}{23!\cdot 2}+\frac{24!\cdot 25}{24!}=300+25=325}\)
1.
\(\displaystyle{ \frac{(2n-1)!\cdot (n+1)!}{n!\cdot (2n+2)!}=\frac{(2n-1)!\cdot n!\cdot (n+1)}{n!\cdot (2n-1)!\cdot 2n (2n+1)\cdot (2n+2)}=\frac{n+1}{2n\cdot 2(n+1)}=\frac{1}{4n}}\)
3.
\(\displaystyle{ (x^{2}+\frac{1}{x^{2}})^{5}= (x^{2})^{5}+{5\choose 1}(x^{2})^{4}\cdot \frac{1}{x^{2}}+{5\choose 2}(x^{2})^{3}\cdot (\frac{1}{x^{2}})^{2})+{5\choose 3}(x^{2})^{2}\cdot (\frac{1}{x^{2}})^{3}+{5\choose 4}x^{2}\cdot (\frac{1}{x^{2}})^{4}+(\frac{1}{x^{2}})^{5}=x^{10}+\frac{5x^{8}}{x^{2}}+\frac{10x^{6}}{x^{4}}+\frac{10x^{4}}{x^{6}}+\frac{5x^{2}}{x^{8}}+\frac{1}{x^{10}}=x^{10}+5x^{6}+10x^{2}+10x^{-2}+5x^{-6}+x^{-10}}\)
\(\displaystyle{ {25\choose 23} +{25\choose 24}=\frac{25!}{23!\cdot 2!}+\frac{25!}{24!}=\frac{23!\cdot 24\cdot 25}{23!\cdot 2}+\frac{24!\cdot 25}{24!}=300+25=325}\)
1.
\(\displaystyle{ \frac{(2n-1)!\cdot (n+1)!}{n!\cdot (2n+2)!}=\frac{(2n-1)!\cdot n!\cdot (n+1)}{n!\cdot (2n-1)!\cdot 2n (2n+1)\cdot (2n+2)}=\frac{n+1}{2n\cdot 2(n+1)}=\frac{1}{4n}}\)
3.
\(\displaystyle{ (x^{2}+\frac{1}{x^{2}})^{5}= (x^{2})^{5}+{5\choose 1}(x^{2})^{4}\cdot \frac{1}{x^{2}}+{5\choose 2}(x^{2})^{3}\cdot (\frac{1}{x^{2}})^{2})+{5\choose 3}(x^{2})^{2}\cdot (\frac{1}{x^{2}})^{3}+{5\choose 4}x^{2}\cdot (\frac{1}{x^{2}})^{4}+(\frac{1}{x^{2}})^{5}=x^{10}+\frac{5x^{8}}{x^{2}}+\frac{10x^{6}}{x^{4}}+\frac{10x^{4}}{x^{6}}+\frac{5x^{2}}{x^{8}}+\frac{1}{x^{10}}=x^{10}+5x^{6}+10x^{2}+10x^{-2}+5x^{-6}+x^{-10}}\)