题目
用数学归纳法证明1+4+9+...+n²=1/6n(n+1)(2n+1)
提问时间:2020-09-28
答案
n=1,成立
n=k,1+4+9+ +k^2=(1/6)k(k+1)(2k+1)
n=k+1,1+4+9+ +k^2+(k+1)^2=(1/6)k(k+1)(2K+1)+(k+1)^2
=[(k+1)(6k+6+2k^2+k)]/6
=[(k+1)(k+2)(2k+2+1)]/6
n=k,1+4+9+ +k^2=(1/6)k(k+1)(2k+1)
n=k+1,1+4+9+ +k^2+(k+1)^2=(1/6)k(k+1)(2K+1)+(k+1)^2
=[(k+1)(6k+6+2k^2+k)]/6
=[(k+1)(k+2)(2k+2+1)]/6
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