Explicație pas cu pas:
cunoaștem formulele:
[tex]\boxed {{1}^{2} + {2}^{2} + {3}^{2} + ... + {n}^{2} = \dfrac{n \cdot (n + 1) \cdot (2n + 1)}{6}}[/tex]
și suma Gauss:
[tex]\boxed {1 + 2 + 3 + ... + n = \dfrac{n \cdot (n + 1)}{2}}[/tex]
atunci limita se scrie:
[tex]= lim_{x\rightarrow + \infty }\dfrac{n \cdot \dfrac{n \cdot (n + 1) \cdot (2n + 1)}{6}}{ {\Big(\dfrac{n \cdot (n + 1)}{2}\Big)}^{2} } = [/tex]
[tex]= lim_{x\rightarrow + \infty } \dfrac{ 4 \cdot {n}^{2} \cdot (n + 1) \cdot (2n + 1)}{6 \cdot {n}^{2} \cdot {(n + 1)}^{2} } \\ [/tex]
[tex]= lim_{x\rightarrow + \infty } = \dfrac{ 4 \cdot (2n + 1)}{6 \cdot (n + 1)} = lim_{x\rightarrow + \infty } = \dfrac{4n + 2}{3n + 3} \\ [/tex]
[tex]= lim_{x\rightarrow + \infty } = \dfrac{n \Big(4 + \dfrac{3}{n} \Big)}{n\Big(3 + \dfrac{3}{n} \Big)} = \bf \dfrac{4}{3} \\ [/tex]
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Explicație pas cu pas:
cunoaștem formulele:
[tex]\boxed {{1}^{2} + {2}^{2} + {3}^{2} + ... + {n}^{2} = \dfrac{n \cdot (n + 1) \cdot (2n + 1)}{6}}[/tex]
și suma Gauss:
[tex]\boxed {1 + 2 + 3 + ... + n = \dfrac{n \cdot (n + 1)}{2}}[/tex]
atunci limita se scrie:
[tex]= lim_{x\rightarrow + \infty }\dfrac{n \cdot \dfrac{n \cdot (n + 1) \cdot (2n + 1)}{6}}{ {\Big(\dfrac{n \cdot (n + 1)}{2}\Big)}^{2} } = [/tex]
[tex]= lim_{x\rightarrow + \infty } \dfrac{ 4 \cdot {n}^{2} \cdot (n + 1) \cdot (2n + 1)}{6 \cdot {n}^{2} \cdot {(n + 1)}^{2} } \\ [/tex]
[tex]= lim_{x\rightarrow + \infty } = \dfrac{ 4 \cdot (2n + 1)}{6 \cdot (n + 1)} = lim_{x\rightarrow + \infty } = \dfrac{4n + 2}{3n + 3} \\ [/tex]
[tex]= lim_{x\rightarrow + \infty } = \dfrac{n \Big(4 + \dfrac{3}{n} \Big)}{n\Big(3 + \dfrac{3}{n} \Big)} = \bf \dfrac{4}{3} \\ [/tex]