Question

1²+2²+...+n²=n(n+1)(2n+1)/6

Answer 1

an
= n^2
=n(n+1) - n
=(1/3)[ n(n+1)(n+2) -(n-1)n(n+1)] - (1/2)[n(n+1) - (n-1)n]
a1+a2+...+an
=(1/3)n(n+1)(n+2) -(1/2)n(n+1)
=(1/6)n(n+1)( 2(n+2) - 3)
=(1/6)n(n+1)( 2n+1)

Answer 2

兄弟这是要干嘛？求证？

Answer 3

这个可以积分求极限证

Answer 4

数学归纳法么:
证明：
（1）当n=1时，左边=1²=1，右边=1×2×3/6=1，等式成立，
（2）假设当n=k时（k为正整数）等式成立，即1²+2²+3²+......+k²=k(k+1)(2k+1)/6，
则当n=k+1时,
左边=1²+2²+3²+......+k²+(k+1)²
=k(k+1)(2k+1)/6+(k+1)²
=[k(k+1)(2k+1)+6(k+1)²]/6
=(k+1)[k(2k+1)+6(k+1)]/6
=(k+1)[2k²+7k+6)]/6
=(k+1)(k+2)(2k+3)/6
=(k+1)(k+2)[2(k+1)+1]/6，
右边=n(n+1)(2n+1)/6=(k+1)(k+2)[2(k+1)+1]/6，也成立，
综上可知:
不论n为何正整时, 1²+2²+3²+......+n²=n(n+1)(2n+1)/6均成立，
得证。