Question

找规律数学题

Answer 1

令S(n)=1²+2²+3²+...+n²，则有S(n)=n(n+1)(2n+1)/6

把n=33带入S(n)=n(n+1)(2n+1)/6可得S(33)=12529

我给你3中推导方法

1.数学归纳法

当n=1时S=1
S(1)=1×(1+1)×(2×1+1)/6=(1×2×3)/6=1，所以n=1时S(n)=n(n+1)(2n+1)/6成立

当n>1且n是正整数时
,假设S(n)=n(n+1)(2n+1)/6,那么就一定有

S(n+1)=(n+1)[(n+1)+1][2(n+1)+1]/6=(n+1)(n+2)(2n+3)/6成立

证:因为S(n+1)

=S(n)+(n+1)²

=[n(n+1)(2n+1)/6]+(n+1)²

=[n(n+1)(2n+1)+6(n+1)²]/6

=(n+1)[(n(2n+1)+6(n+1)]/6

=(n+1)(2n²+7n+6)/6

=(n+1)(n+2)(2n+3)/6

所以S(n)=n(n+1)(2n+1)/6成立

2.立方和公式证法（n+1)³=n³+3n²+3n+1
所以

（n+1)³-n³=3n²+3n+1

n³-(n-1)³=3(n-1)²+3(n-1)+1

.
..
..
.
3³-2³=3×2²+3×2+1
2³-1³=3×1²+3×1+1
左右相加
（n+1)³-1=3×(1²+2²+3²+...+n²)
+3×n(n+1)/2+n
化剑可得得到
S=1²+2²+3²+...+n²=n(n+1)(2n+1)/6

3.记An=1²+2²+3²+n²1²+2²+3²+n²-[n(n+1)(2n+1)/6]

那么A(n+1)-An

=1²+2²+3²+n²+(n+1)²-[(n+1)(n+2)(2n+3)/6]-1²+2²+3²+n²+[n(n+1)(2n+1)/6]

=(n+1)²--[(n+1)(n+2)(2n+3)/6]+[n(n+1)(2n+1)/6]

=(n+1)[(n+1)-(n+2)(2n+3)/6+n(2n+1)/6]

=(n+1)[(n+1)-(n+1)]=0

因为A1=1²-1×(1+2)(2×1+1)/6=0，所以An=0

所以1²+2²+3²+n²1²+2²+3²+n²-[n(n+1)(2n+1)/6]=0

所以S=1²+2²+3²+n²1²+2²+3²+n²=[n(n+1)(2n+1)/6]

Answer 2

规律是2^n-1
第2013个数应该是
2^2012