数学

学术讨论会

那就浅看一下大家技术怎么样

$\sum_{i=1}^n\sum\limits_{j=i}n(i+j)$

$\sum_{i=1}^{n} \sum_{j=i}^{n} (i + j)$

$\sum_{i=1}^{n} \sum_{j=i}^{n} i$

$\sum_{j=i}^{n} i = i \cdot (n - i + 1)$

$\sum_{i=1}^{n} i (n - i + 1) = \sum_{i=1}^{n} [i(n + 1) - i^2] = (n + 1) \sum_{i=1}^{n} i - \sum_{i=1}^{n} i^2$

$代入公式：\sum_{i=1}^{n} i = \frac{n(n+1)}{2}, \quad \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$

$得：(n + 1) \cdot \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)^2}{2} - \frac{n(n+1)(2n+1)}{6}$

$n(n+1) \left[ \frac{n+1}{2} -\frac{2n+1}{6} \right]=n(n+1)\left[\frac{3(n+1)-(2n+1)}{6}\right]=n(n+1)\cdot\frac{n+2}{6}=\frac{n(n+1)(n+2)}{6}\]$ $\sum_{i=1}^{n} \sum_{j=i}^{n} j\)$

$\sum_{j=i}^{n} j = \sum_{j=1}^{n} j - \sum_{j=1}^{i-1} j = \frac{n(n+1)}{2} - \frac{(i-1)i}{2} = \frac{n(n+1) - i(i-1)}{2}$

$\sum_{i=1}^{n} \frac{n(n+1) - i(i-1)}{2} = \frac{1}{2} \sum_{i=1}^{n} \left[ n(n+1) - (i^2 - i) \right]$

$\sum_{i=1}^{n} \left[ n(n+1) - (i^2 - i)\right] = \sum_{i=1}^{n} n(n+1) -\sum_{i=1}^{n} (i^2-i)=n^2(n+1)-\left(\sum_{i=1}^{n}i^2- \sum_{i=1}^{n}i\right)$

$代入公式：\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{i=1}^{n} i = \frac{n(n+1)}{2}$

$得：n^2 (n+1) - \left( \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} \right) = n^2 (n+1) - \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

$n(n+1)\left[n-\frac{2n+1}{6}+\frac{1}{2}\right]=n(n+1)\left[\frac{6n-(2n+1)+3}{6}\right]=n(n+1)\cdot\frac{4n+2}{6}=n(n+1)\cdot\frac{2(2n+1)}{6}=\frac{n(n+1)(2n+1)}{3}$

$因此：\frac{1}{2} \cdot \frac{n(n+1)(2n+1)}{3} = \frac{n(n+1)(2n+1)}{6}$

$\frac{n(n+1)(n+2)}{6} + \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{6} \left[ (n+2) + (2n+1) \right] $

$= \frac{n(n+1)}{6} (3n + 3) = \frac{n(n+1)}{6} \cdot 3(n+1) = \frac{3n(n+1)^2}{6} = \frac{n(n+1)^2}{2}$

$结果为\frac{n(n+1)^2}{2}\).\[\boxed{\dfrac{n(n+1)^{2}}{2}}$