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#15939: 利用公式計算


tang891228 (tang891228)

a216. 數數愛明明 | From: [1.172.99.56] | 發表日期: 2018-11-07 03:05

公式推導：

{\displaystyle {\begin{aligned}f(n)&=1+2+\cdots +n\\&=\sum \limits _{k=1}^{n}k\\&={\frac {n(n+1)}{2}}\end{aligned}}}

 

{\displaystyle {\begin{aligned}g(n)&=f(1)+f(2)+\cdots +f(n)\\&=\sum \limits _{k=1}^{n}{\frac {k(k+1)}{2}}\\&={\frac {1}{2}}\cdot {\frac {n(n+1)(2n+1)}{6}}+{\frac {1}{2}}\cdot {\frac {n(n+1)}{2}}\\&={\frac {n(n+1)(n+2)}{6}}\end{aligned}}}

#20844: Re:利用公式計算


lucianuschen@gmail.com (L Ch)

a216. 數數愛明明 | From: [218.32.127.71] | 發表日期: 2020-03-12 10:35

公式推導：

{\displaystyle {\begin{aligned}f(n)&=1+2+\cdots +n\\&=\sum \limits _{k=1}^{n}k\\&={\frac {n(n+1)}{2}}\end{aligned}}}

 

{\displaystyle {\begin{aligned}g(n)&=f(1)+f(2)+\cdots +f(n)\\&=\sum \limits _{k=1}^{n}{\frac {k(k+1)}{2}}\\&={\frac {1}{2}}\cdot {\frac {n(n+1)(2n+1)}{6}}+{\frac {1}{2}}\cdot {\frac {n(n+1)}{2}}\\&={\frac {n(n+1)(n+2)}{6}}\end{aligned}}}

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