- math求证:1^4+2^4+3^4+……+n^4=n(n+1)
- 求证:1^4+2^4+3^4+……+n^4=n(n+1)(2n+1)(3n^2+3n-1)/30
- 归纳法：
n=1时,左边=右边
若n=k时等式成立,则1^4+2^4+3^4+……+k^4=k(k+1)(2k+1)(3k^2+3k-1)/30
所以 1^4+2^4+3^4+……+k^4+(k+1)^4=k(k+1)(2k+1)(3k^2+3k-1)/30+(k+1)^4,而(k+1)(k+1+1)(2k+2+1)[3(k+1)^2+3k+3-1)/30-k(k+1)(2k+1)(3k^2+3k-1)/30=(k+1)(30k^3+90k^2+90k+30)/30=(k+1)^4
所以1^4+2^4+3^4+……+k^4+(k+1)^4=(k+1)(k+1+1)(2k+2+1)[3(k+1)^2+3k+3-1)/30
所以若n=k成立,则n=k+1成立
所以一切正整数n均使等式成立