找规律数学题
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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]
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]
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