Răspuns:
Explicație pas cu pas:
[tex]\sqrt[3]{1\times2\times3} =\sqrt[3]{6} < \sqrt[3]{8} =2\\\sqrt[3]{2\times3\times4} =\sqrt[3]{24} < \sqrt[3]{27} =3\\.\\.\\.\\\sqrt[3]{(n-1)\times n\times(n+1)} =\sqrt[3]{n^{3} -n} < \sqrt[3]{n^{3}} =n\\[/tex]
[tex]\sqrt[3]{1\times 2\times3}+\sqrt[3]{2\times 3\times 4}+...+\sqrt[3]{(n-1)\times n\times(n+1)} < 2+3+...+n < 1+2+3+..+n[/tex]
Știm:
[tex]1+2+3+...+n=\frac{n\times(n+1)}{2}= \frac{n^{2}+n}{2}[/tex]
[tex]\sqrt[3]{1\times 2\times3}+\sqrt[3]{2\times 3\times 4}+...+\sqrt[3]{(n-1)\times n\times(n+1)} < \frac{n^{2}+n }{2} < \frac{n^{2}+2n }{2}[/tex]
=>[tex]\sqrt[3]{1\times 2\times3}+\sqrt[3]{2\times 3\times 4}+...+\sqrt[3]{(n-1)\times n\times(n+1)} < \frac{n^{2}+2n }{2}[/tex]
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Răspuns:
Explicație pas cu pas:
[tex]\sqrt[3]{1\times2\times3} =\sqrt[3]{6} < \sqrt[3]{8} =2\\\sqrt[3]{2\times3\times4} =\sqrt[3]{24} < \sqrt[3]{27} =3\\.\\.\\.\\\sqrt[3]{(n-1)\times n\times(n+1)} =\sqrt[3]{n^{3} -n} < \sqrt[3]{n^{3}} =n\\[/tex]
[tex]\sqrt[3]{1\times 2\times3}+\sqrt[3]{2\times 3\times 4}+...+\sqrt[3]{(n-1)\times n\times(n+1)} < 2+3+...+n < 1+2+3+..+n[/tex]
Știm:
[tex]1+2+3+...+n=\frac{n\times(n+1)}{2}= \frac{n^{2}+n}{2}[/tex]
[tex]\sqrt[3]{1\times 2\times3}+\sqrt[3]{2\times 3\times 4}+...+\sqrt[3]{(n-1)\times n\times(n+1)} < \frac{n^{2}+n }{2} < \frac{n^{2}+2n }{2}[/tex]
=>[tex]\sqrt[3]{1\times 2\times3}+\sqrt[3]{2\times 3\times 4}+...+\sqrt[3]{(n-1)\times n\times(n+1)} < \frac{n^{2}+2n }{2}[/tex]