∑の計算

$\sum\limits_{k = 1}^n 1 = n$

$\sum\limits_{k = 1}^n k = \frac{{n\left( {n + 1} \right)}}{2}$

$\sum\limits_{k = 1}^n {k^2 } =$ $\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$

$\sum\limits_{k = 1}^n {k^3 } = \frac{{n^2 \left( {n + 1} \right)^2 }}{4}$

[し]
[さ行]
[索引トップ]

by数学ナビゲーター