1+2+3+4+......+121=?的计算公式

2025年03月15日 15:00
有5个网友回答
网友(1):

令Sn=1+2+3+4+......+n
则Sn=n+(n-1)+(n-2)+(n-3)+......+1
则两个等式左右相加得2Sn=(n+1)+(n+1)+(n+1)+(n+1)......(n+1)=(n+1)*n
则Sn=(n+1)*n/2(即高中的等差数列求和公式,很简单的,相信参加过高考的,都会记得)

代入n=121
可得Sn=7381(即为所求等的差数列和)

网友(2):

这个是小学就会的求和公式吧,(首项+尾项)*项数/2
Sn=(1+121)*121/2=7381

网友(3):

写出原来的计算式,在原式的下面再以相反的顺序写一式:

1+2+3+4+......+121=?
121+122+......+4+3+2+1=?

上面一行的第1项与下面一行的第1项相加,第2项与第2项相加,第3项与与第3项相加......,第121项与第121项相加可得:
(121+1)*121=2*?
得:?=61*121=7381

网友(4):

(1+120)*60+121
就可以了

网友(5):

(1+121)/2*121

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