Matematică
ovidiupaula
2015-11-16 06:14:39
să se calculeze radical din A, daca:
Răspunsuri la întrebare
ioaa1
2015-11-16 08:20:30

[latex]A = dfrac{1}{1cdot4}+dfrac{1}{2cdot 6}+dfrac{1}{3cdot 8}+dfrac{1}{4cdot 10}+...+dfrac{1}{48cdot 98}+dfrac{1}{49cdot 100} \ \ A = sumlimits_{k=1}^{49}dfrac{1}{kcdot (2k+2)} = sumlimits_{k=1}^{49}dfrac{1}{kcdot 2(k+1)} = sumlimits_{k=1}^{49}dfrac{1}{2}cdotdfrac{1}{kcdot (k+1)} = \ \ [/latex] [latex]=dfrac{1}{2}cdot sumlimits_{k=1}^{49}dfrac{1}{kcdot (k+1)} = dfrac{1}{2}cdot sumlimits_{k=1}^{49}Big(dfrac{1}{k}-dfrac{1}{k+1}Big)}= \ \ =dfrac{1}{2}cdotBig(sumlimits_{k=1}^{49}dfrac{1}{k}-sumlimits_{k=1}^{49}dfrac{1}{k+1}Big) = \ \ = dfrac{1}{2}cdot Big(dfrac{1}{1}+dfrac{1}{2}+dfrac{1}{3}+...+dfrac{1}{49}-Big(dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+...+dfrac{1}{50}Big)Big) =[/latex] [latex]= dfrac{1}{2}cdotBig(1+dfrac{1}{2}+dfrac{1}{3}+...+dfrac{1}{49}-Big(dfrac{1}{2}+dfrac{1}{3}+ dfrac{1}{4}+...+dfrac{1}{49}Big)-dfrac{1}{50}Big) = \ \ = dfrac{1}{2}cdot Big(1-dfrac{1}{50}Big) = dfrac{1}{2}cdot Big(dfrac{50-1}{50} Big)= dfrac{1}{2}cdot dfrac{49}{50} = dfrac{49}{100} \ \ Rightarrow sqrt{A} = sqrt{dfrac{49}{100}}= dfrac{7}{10}[/latex] [latex]\ $M-am folosit de proprietatile sumei (Sigma): \ \ left| egin{array}{c} sumlimits_{ig{k=1}}^{ig{n}}acdot k =acdot sumlimits_{ig{k=1}}^{ig{n}} k quadquad quad quad $ $ \ sumlimits_{ig{k=1}}^{ig{n}}ig(a+big) = sumlimits_{ig{k=1}}^{ig{n}}a+sumlimits_{ig{k=1}}^{ig{n}}bend{array} ight |[/latex] Cu suma (Sigma) se rezolva foarte lejer.

Mizukamaru
2015-11-16 08:21:45

[latex]it dfrac{1}{1cdot4} +dfrac{1}{2cdot6} +dfrac{1}{3cdot8} + ... dfrac{1}{49cdot100} = \;\ \;\ =dfrac{1}{2} left(dfrac{1}{1cdot2} +dfrac{1}{2cdot3} +dfrac{1}{3cdot4} + ... dfrac{1}{49cdot50} ight) [/latex] Fiecare fracție din paranteză se descompune după formula: [latex]it dfrac{1}{k(k+1)} =dfrac{1}{k}-dfrac{1}{k+1}[/latex] Prin urmare, vom avea: [latex]it dfrac{1}{2} left(dfrac{1}{1}-dfrac{1}{2} +dfrac{1}{2}-dfrac{1}{3} +dfrac{1}{3} -dfrac{1}{4}+ ... dfrac{1}{49} -dfrac{1}{50} ight) =dfrac{1}{2}left(dfrac{1}{1}-dfrac{1}{50} ight)= \;\ \;\ =dfrac{1}{2}cdotdfrac{49}{50} =dfrac{49}{100}[/latex] Se cere radical din acest ultim rezultat: [latex]it sqrt{dfrac{49}{100}} =dfrac{sqrt{49}}{sqrt{100}} =dfrac{7}{10}.[/latex]

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