用数学归纳法证明1的2次方+2的2次方+……+n的2次方=n(n+1)(2n

Question

用数学归纳法证明1的2次方+2的2次方+……+n的2次方=n(n+1)(2n

Answer 1

结果应该是n(n+1)(2n+1)/6 吧?

数学归纳法
1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6
当n=1时,显然成立.

设n=k时也成立,即:
1^2+2^2+3^2+……+k^2=k(k+1)(2k+1)/6

那么当n=k+1时,等式的左边等于:
1^2+2^2+3^2+……+k^2+(k+1)^2
=k(k+1)(2k+1)/6+(k+1)^2
=(k+1)[k(2k+1)/6+(k+1)]
=(k+1)[2k^2+k+6k+6]/6
=(k+1)(k+2)(2k+3)/6

而等式的右边等于:(当n=k+1时)
(k+1)(k+1+1)(2k+2+1)/6
=(k+1)(k+2)(2k+3)/6

即当n=k+1时,等式左边等于等式的右边
所以对于一切n,等式都成立