Ci膮gi, zadanie nr 4875
ostatnie wiadomo艣ci | regulamin | latex
| Autor | Zadanie / Rozwi膮zanie |
fazi post贸w: 26 |  2015-01-08 11:29:51oblicz granic臋 ci膮gu o wyrazie og贸lnym: prosz臋 rozpisa膰 a)$a_{n}=\frac{2+4+6+...+2n}{n^{2}}$ b)$a_{n}=\frac{1+3+5+...+(2n-1)}{n(n+1)}$ c)$a_{n}=\frac{1+2+3+...+n}{n^{2}+2n+5}$ d)$a_{n}=3n-\sqrt{9n^{2}+6n-5}$ e)$a_{n}=\sqrt{n^{2}-1}-\sqrt{n^{2}-2}$ f)$a_{n}=\sqrt{4n^{2}+n-2n}$ |
irena post贸w: 2636 | 2015-01-08 11:42:45 $a_n=\frac{2+4+6+...+2n}{n^2}$ $2+4+6+...+2n=\frac{2+n}{2}\cdot n=\frac{2n+n^2}{2}$ $a_n=\frac{2n+n^2}{2n^2}=\frac{\frac{2}{n}+1}{2}\to\frac{1}{2}$ |
irena post贸w: 2636 | 2015-01-08 11:45:11 $a_n=\frac{1+3+5+...+2n-1}{n(n+1)}$ $1+3+5+...+2n-1=\frac{1+2n-1}{2}\cdot n=n^2$ $a_n=\frac{n^2}{n^2+n}=\frac{1}{1+\frac{1}{n}}\to1$ |
irena post贸w: 2636 | 2015-01-08 11:47:28 $a_n=\frac{1+2+3+...+n}{n^2+2n+5}$ $1+2+3+...+n=\frac{1+n}{2}\cdot n=\frac{n(n+1)}{2}$ $a_n=\frac{n^2+n}{2n^2+4n+10}=\frac{1+\frac{1}{n}}{2+\frac{4}{n}+\frac{10}{n^2}}\to\frac{1}{2}$ |
irena post贸w: 2636 | 2015-01-08 11:51:09 $a_n=3n-\sqrt{9n^2+6n-5}=\frac{9n^2-9n^2-6n+5}{3n+\sqrt{9n^2+6n-5}}=$ $=\frac{-6n+5}{3n+n\sqrt{9+\frac{6}{n}-\frac{5}{n^2}}}=$ $=\frac{-6+\frac{5}{n}}{3+\sqrt{9+\frac{6}{n}-\frac{5}{n^2}}}\to\frac{-6}{3+3}=-1$ |
irena post贸w: 2636 | 2015-01-08 11:54:31 $a_n=\sqrt{n^2-1}-\sqrt{n^2-2}=\frac{n^2-1-n^2+2}{\sqrt{n^2-1}+\sqrt{n^2-2}}=\frac{1}{n\sqrt{1-\frac{1}{n^2}}+n\sqrt{1-\frac{2}{n^2}}}=$ $=\frac{\frac{1}{n}}{\sqrt{1-\frac{1}{n^2}}+\sqrt{1-\frac{2}{n^2}}}\to\frac{0}{1+1}=0$ |
irena post贸w: 2636 | 2015-01-08 11:58:09 Chyba powinno by膰; $a_n=\sqrt{4n^2+n}-2n=\frac{4n^2+n-4n^2}{\sqrt{4n^2+n}+2n}=\frac{n}{n\sqrt{4+\frac{1}{n}}+2n}=$ $=\frac{1}{\sqrt{4+\frac{1}{n}}+2}\to\frac{1}{2+2}=\frac{1}{4}$ |
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