YLL討論網

#ed_op#DIV#ed_cl#利用下一個平方數會是前一個平方數的:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(n+1)#ed_op#SUP#ed_cl#2#ed_op#SUB#ed_cl#=n#ed_op#/SUB#ed_cl#2 #ed_op#SUB#ed_cl#+ n + (n+1) #ed_op#/SUB#ed_cl##ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SUB#ed_cl##ed_op#SUP#ed_cl#以及產生平方數的底數(index)是奇數, 1, 3, 5, 7, 9,.......#ed_op#/SUP#ed_cl##ed_op#/SUB#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#假設k是奇數且a#ed_op#SUB#ed_cl#k#ed_op#/SUB#ed_cl#=n#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a#ed_op#SUB#ed_cl#(k+1)  #ed_op#/SUB#ed_cl##ed_op#SUP#ed_cl##ed_op#SUB#ed_cl#= n#ed_op#/SUB#ed_cl#2#ed_op#SUB#ed_cl#  +   n;#ed_op#/SUB#ed_cl##ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SUP#ed_cl##ed_op#SUB#ed_cl#a(k+2) =  ( n#ed_op#/SUB#ed_cl#2#ed_op#SUB#ed_cl# + n  )  + #ed_op#U#ed_cl#(n + 1 )#ed_op#/U#ed_cl#      #ed_op#/SUB#ed_cl##ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SUP#ed_cl##ed_op#SUB#ed_cl#            = (n+1)#ed_op#/SUB#ed_cl# 2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SUP#ed_cl##ed_op#/SUP#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SUP#ed_cl#寫的比較簡略,請見諒#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SUP#ed_cl#太久沒來使用,有點不太會用了#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

考慮此數列a(n). 其定義為:
a(1)=1,.
證明此數列含有無限個完全平方數.
([]表高斯符號)