23+25+27+...+45=?

如果用高斯的猜想方法 (即1+2+3...+99+100=?)
23+25+27+...+45=

(23+45)+(25+43)+...+(33+35) 共有 6組相同的和
= 6x68
= 408

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如果用等差級數來計算的話

25 = 23+2=
27 = 23+2+2 = 23+(2x2)
29 = 23+2+2+2 = 23+(2x3)
...
45 = 23+(2x11)

23+25+27+...+45 = 23+(23+2)+[23+(2x2)]+...+[23+(2x11)]

This is an arithmetic progression. An arithmetic progression is a sequence of numbers in which the difference between two consecutive numbers is always the same.

sum of A.P. = (1/2)(n)(2a+(n−1)d)

這裡的n(總項數)是12, d(共差)是2, a(首數)=23

(1/2)(12)[(2)(23)+(12-1)(2)]
=6(46+22)
=408

23+25+27+...+45=? 計算過程係點?請您教教我了!

[(1+45)/2]^2-[(1+21)/2]^2

將隔數先加(如:23+37=50)就好記D 25系圓整吾好加住 跟住就29+31=60 33+37=70 39+41=80 最後25+35+45=105 60+70+80=310(60拿80ge10就變70 甘3*70=210) 210加反105=315 希望可以解答到你ge問題