幾個級數和的公式
Author: 藍邦偉  老師

幾個級數和的公式

 \sum_{k=1 }^{ n}k=1+2+...+(n-1)+n= \frac{ n(n+1) }{2 }  

 \sum_{k=1 }^{ n}k²=1²+2²+...+(n-1)²+n²= \frac{ n(n+1)(2n+1) }{6 } 

 \sum_{k=1 }^{ n}k³=1³+2³+...+(n-1)³+n³=[ \frac{ n(n+1) }{2 }]² 

公式

1+2+...+n

.

 \sum_{k=1 }^{n } k²

.

 \sum_{k=1 }^{n } k³

 \sum_{k=1 }^{n }k 

n = 68 (slider from 1 to 100, step 1)
L1 = Sequence(k, k, 1, n)
a_1 = Sum(L1)
L2 = Sequence(k^(2), k, 1, n)
a_2 = Sum(L2)
L3 = Sequence(k^(3), k, 1, n)
a_3 = Sum(L3)
文字1: " \sum_{ k=1}^{" + n + " } k=\frac{" + n + "\times{}" + (n + 1) + "  }{ 2} =" + a_1
文字2: "\sum_{k=1 }^{" + n + " }k²= \frac{" + n + "\times{}" + (n + 1) + "\times{}" + (2 * n + 1) + "}{6 }= " + a_2
文字3: " \sum_{k=1 }^{" + n + " }k³=[ \frac{ " + n + "\times{}" + (n + 1) + " }{ 2}]²= " + a_3
文字4: "\sum_{k=1 }^{" + n + " } \frac{ 1 }{ k(k+1)}= \sum_{k=1 }^{" + n + " } ( \frac{ 1 }{ k} - \frac{ 1 }{k+1 })=1- \frac{ 1 }{" + n + " } = \frac{ " + (n - 1) + " }{" + n + " } "
文字9 = If(n > 20, Sum[Sequence[Text["" + (i) + "+"], i, 1, 10]] + "............+" + (Sum[Sequence[Text["" + (i) + (If[i ≟ n, "=" + (n * (n + 1) / 2), "+"])], i, n - 3, n]]), Sum[Sequence[Text["" + (i) + (If[i ≟ n, "=" + (n * (n + 1) / 2), "+"])], i, 1, n]])
a = false
b = false
c = false
文字11 = If(n > 20, Sum[Sequence[Text["" + (i^(2)) + "+"], i, 1, 10]] + "............+" + (Sum[Sequence[Text["" + (i^(2)) + (If[i ≟ n, "=" + (n * (n + 1) * (2 * n + 1) / 6), "+"])], i, n - 3, n]]), Sum[Sequence[Text["" + (i^(2)) + (If[i ≟ n, "=" + (n * (n + 1) * (2 * n + 1) / 6), "+"])], i, 1, n]])
d = false
文字13 = If(n > 20, Sum[Sequence[Text["" + (i^(3)) + "+"], i, 1, 10]] + "............+" + (Sum[Sequence[Text["" + (i^(3)) + (If[i ≟ n, "=" + ((n * (n + 1) / 2)^(2)), "+"])], i, n - 3, n]]), Sum[Sequence[Text["" + (i^(3)) + (If[i ≟ n, "=" + ((n * (n + 1) / 2)^(2)), "+"])], i, 1, n]])