Discuss Scratch

#1Oct. 6, 2018 01:08:56
KeepInventory
Scratcher
31 posts

scratch版极智空间

谁能证明
<[((((((((1) * (1)) * (1))) + ((((2) * (2)) * (2)))) + ((((3) * (3)) * (3)))) + (......)) + ((((n) * (n)) * (n))))] = (((((((n) * (n)) * (n)) * (n)) + (((((2) * (n)) * (n)) * (n)))) + ((n) * (n))) / (4))>




Last edited by KeepInventory (Oct. 6, 2018 01:15:28)

#2Oct. 6, 2018 01:23:18
KeepInventory
Scratcher
31 posts

scratch版极智空间


(((1)*(1))*(1))+(((2)*(2))*(2))+(((3)*(3))*(3))+(......)+((((foo))*(foo))*(foo))=（((((((n)*(n))*(n))*(n))+(((2)*(n))*(n))*(n))+((n)*(n))))∕4)))

Last edited by KeepInventory (Oct. 6, 2018 01:24:03)

#3Oct. 7, 2018 10:32:52
KeepInventory
Scratcher
31 posts

scratch版极智空间

转成scratchblocks格式：
1*1*1+2*2*2+3*3*3+……+n*n*n=（n*n*n*n+2*n*n*n+n*n)/4
把……改成如下形式
(...... :: ......)

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