观察下列各式:
13+23=1+8=9,而(1+2)2=9,∴13+23=(1+2)2;
13+23+33=6,而(1+2+3)2=36,∴13+23+33=(1+2+3)2;
13+23+33+43=100,而(1+2+3+4)2=100,∴13+23+33+43=(1+2+3+4)2;
∴13+23+33+43+53=(________)2=________.
根据以上规律填空:
(1)13+23+33+…+n3=(________)2=[________]2.
(2)猜想:113+123+133+143+153=________.
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解:由题意可知:13+23+33+43+53=(1+2+3+4+5)2=225
(1)∵1+2+…+n=(1+n)+[2+(n-1)]+…+[+(n-+1)]=,
∴13+23+33+…+n3=(1+2+…+n)2=[]2;
(2)113+123+133+143+153=13+23+33+…+153-(13+23+33+…+103)
=(1+2+…+15)2-(1+2+…+10)2
=1202-552=11375.
故