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Cat face radical din 1+2+3+...+100 supra 1+3+5+...+333?

Răspunsuri la întrebare

Răspuns de falcuta205
1
[tex]\sqrt{\frac{1+2+3+...+100}{1+3+5+...+333}}\\S1=1+2+3+...+100\\ S1=100+99+98+...+1\\Adunanad\ membru\ cu\ membru\\ 2S1=101+101+101+...+101(suma\ are\ 100\ termeni)\\2S1=100*101\\ S1=\frac{100*101}{2}=50*101=5050\\S2=1+3+5+...+333\\1=2*0+1\\3=2*1+1\\ 5=2*2+1\\...\\...\\...\\333=2*166+1\\S2=(2*0+1)+(2*1+1)+(2*2+1)+...+(2*166+1) \\S2=2*0+2*1+2*2+....+2*166+1+1+1+...+1\\1\ se\ aduna\ de\ 167\ de\ ori\\ S2=0+2*(1+2+3+...+166)+1*167\\S2=2*\frac{166*167}{2}+167\\ S2=167*166+167*1\\S2=167*(166+1)\\S2=167*167\\S2=167^{2}\\S2=27889\\ \sqrt{\frac{5050}{27889}}=\frac{\sqrt{5050}}{\sqrt{27889}}= \frac{5\sqrt{202}}{167}[/tex]
Răspuns de Utilizator anonim
0
\displaystyle  \sqrt{\frac{1+2+3+...+100}{1+3+5+...+333}}  \\  \\ 1+2+3+...+100= \frac{100(100+1)}{2} = \frac{100 \cdot 101}{2} =50 \cdot 101=5050  \\  \\ 1+2+3+...+100=5050

\displaystyle 1+3+5+...+333 \\ \\ r=3-1 \Rightarrow r=2 \\ \\333=1+(n-1) \cdot 2 \Rightarrow 333=1+2n-2 \Rightarrow 2n=333-1+2 \Rightarrow  \\ \\ \Rightarrow 2n=334 \Rightarrow n= \frac{334}{2} \Rightarrow n=167 \\  \\ S_{167}= \frac{2 \cdot 1+(167-1) \cdot 2}{2} \cdot 167= \frac{2+166 \cdot 2}{2} \cdot 167= \\  \\ = \frac{2+332}{2} \cdot 167= \frac{334}{2} \cdot 167=167 \cdot 167=27889

\displaystyle 1+3+5+...+333=27889 \\ \\  \sqrt{ \frac{1+3+3+...+100}{1+3+5+...+333} } = \sqrt{ \frac{5050}{27889} } = \frac{ \sqrt{5050} }{ \sqrt{27889} }= \frac{5 \sqrt{202} }{167}

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