1²+2²+3²+…+（n-1)²=n(n-1)(2n-1)/6
问题描述：
1²+2²+3²+…+（n-1)²=n(n-1)(2n-1)/6
通俗易懂些,本人还只是初三,作业里面有这题
答
By M.I.
n=1
LS = 0^2 =0
RS =0
n=1 is true
Assum n=k is true
ie 1²+2²+3²+…+（k-1)²=k(k-1)(2k-1)/6
for n=k+1
LS =1²+2²+3²+…+（k-1)²+k²
= k(k-1)(2k-1)/6 +k²
= k{ [(k-1)(2k-1) +6k] /6}
= k{ (2k^2+3k+1)/6}
= k(2k+1)(k+1)/6
= (k+1)(k+1-1)(2(k+1)-1)/6 = RS
By principle of MI,it is true for all n