用数学归纳法证明:1*(n^2-1^2)+2*(n^2-2^2)...+n(n^2-n^2)=1/4n^4-1/4n^2
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用数学归纳法证明:1*(n^2-1^2)+2*(n^2-2^2)...+n(n^2-n^2)=1/4n^4-1/4n^2
用数学归纳法证明:1*(n^2-1^2)+2*(n^2-2^2)...+n(n^2-n^2)=1/4n^4-1/4n^2
用数学归纳法证明:1*(n^2-1^2)+2*(n^2-2^2)...+n(n^2-n^2)=1/4n^4-1/4n^2
当n=1
左边=0
右边=0
成立
设n=k时也成立
1*(k^2-1^2)+2*(k^2-2^2)...+k(k^2-k^2)=1/4k^4-1/4k^2
当n=k+1时
左边=1*((k+1)^2-1^2)+2*((k+1)^2-2^2)...+k((k+1)^2-k^2)-(k+1)((k+1)^2-(k+1)^2)
=1*(k^2-1^2)+2*(k^2-2^2)...+k(k^2-k^2)+(2k+1)(1+2+3...k)
=1/4k^4-1/4k^2+(2k+1)k(k+1)/2
=1/4(k+1)^4-1/4(k+1)^2
得证