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1、求1+2+3+4+5+6+…+n的公式是一个非常有名的等差数列求和问题,从第二项起,每一项与它的前一项的差等于同一个常数的一种数列,常用A、P表示。

缘由无非是这两个视频给出了两个看似匪夷所思的等式。 第一期告诉我们格兰迪 (Grandi)级数等于1/2。 第二期的结论似乎更惊悚,全体自然数之和等于-1/12! 怎么可能? That is impossible! 相信这是大多数人的第一反应。 第一个4出现之前已经有了一个1两个2三个3,所以第一个4出现在n=1+2+3+1的位置。 第一个n出现之前已经有了一个1两个2三个3四个4...,所以第一个n出现在n=1+2+3+...(n-1)+1的位置上。 1 人赞同了该回答 看了前面五个人的回答,我也有一个答案和它们不一样: 1+2=3=1 \times 2+1; 2+3=8=2 \times 3+2; 3+4=15=3 \times 4+3; 4+5=24=4 \times\ 5+4; 6+7=6 \times 7+6=48。 说明每一项与后一项相差2n+2,可知它们每一项之差形成 等差数列,所以这个题目是个二级等差数列求和 而二级等差数列的前n项和满足 二次函数

能否设计一个连续函数,使f (1,2,3,4)=1,2,3,4。 f (5)=114514? 时隔近五年的更新,问题终结,不用回答了。 题主在问这个问题的大约一年后,就学了数值计算方法 显示全部 关注者 43 333300。 计算过程如下: 1×2+2×3+3×4+4×5+5×6+…+99×100 =1×(1+1)+2×(2+1)+3×(3+1)+…+98×(98+1)+99×(99+1) =12+1+22+2+32+3+…+992. 每一个计算式的正常计算结果+前一个计算式的结果就是答案:比如“2+3=”正常计算结果5+前一个计算式结果3就=8;“3+4=”正常计算结果为7,加上前一计算式结果8就=15;所以先计算“5+6”规律结果应为11+24=35,再来计算“6+7”,结果就是13+35=48。 不知道对不对,因为如果要找规律,除了我找的. 两边求和,我们有 ln (n+1)<1/1+1/2+1/3+1/4+……+1/n 容易的, \lim _ {n\rightarrow +\infty }\ln \left ( n+1\right) =+\infty ,所以这个和是无界的,不收敛。

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