平方数列、立方数列之求和
平方和
1 2 + 2 2 + ⋯ + n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}6 12+22+⋯+n2=6n(n+1)(2n+1)
推导：（利用立方差公式）
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n^3-(n-1)^3=1\times[n^2+n(n-1)+(n-1)^2]=2n^2+(n-1)^2-n
n3−(n−1)3=1×[n2+n(n−1)+(n−1)2]=2n2+(n−1)2−n
由此得到
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\begin{aligned} 2^3-1^3&=2\times2^2+1^2-2\\ 3^3-2^3&=2\times3^2+2^2-3\\ &\vdots\\ n^3-(n-1)^3&=2n^2+(n-1)^2-n\\ \end{aligned}
23−1333−23n3−(n−1)3=2×22+12−2=2×32+22−3⋮=2n2+(n−1)2−n
上式两边分别相加得到：
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I=1^2+2^2+\cdots+n^2
I=12+22+⋯+n2）
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\begin{aligned} n^3-1^3 &=2(2^2+3^2+\cdots+n^2)+[1^2+2^2+\cdots+(n-1)^2]-(2+3+\cdots+n)\\ &=2I-2+I-n^2-\frac{n(1+n)}{2}+1\\ &=3I-1-\frac32n^2-\frac n2\\ &=3I-\frac{3n^2+n+2}{2} \end{aligned}
n3−13=2(22+32+⋯+n2)+[12+22+⋯+(n−1)2]−(2+3+⋯+n)=2I−2+I−n2−2n(1+n)+1=3I−1−23n2−2n=3I−23n2+n+2
于是
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3I=n^3+\frac{3n^2+n}2\Longrightarrow I=\frac{2n^3+3n^2+n}6=\frac{n(n+1)(2n+1)}6.
3I=n3+23n2+n⟹I=62n3+3n2+n=6n(n+1)(2n+1).
立方和
1 3 + 2 3 + ⋯ + n 3 = [ n ( n + 1 ) 2 ] 2 = n 4 + 2 n 3 + n 2 4 1^3+2^3+\cdots+n^3=\left[\frac{n(n+1)}{2}\right]^2=\frac{n^4+2n^3+n^2}4 13+23+⋯+n3=[2n(n+1)]2=4n4+2n3+n2
同样地，根据
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n^4-(n-1)^4=[n^2+(n-1)^2]\times[n^2-(n-1)^2]=4n^3-6n^2+4n-1
n4−(n−1)4=[n2+(n−1)2]×[n2−(n−1)2]=4n3−6n2+4n−1
得到
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\begin{aligned} 2^4-1^4&=4\cdot2^3-6\cdot2^2+4\cdot2-1\\ 3^4-2^4&=4\cdot3^3-6\cdot3^2+4\cdot3-1\\ &\vdots\\ n^4-(n-1)^4&=4n^3-6n^2+4n-1 \end{aligned}
24−1434−24n4−(n−1)4=4⋅23−6⋅22+4⋅2−1=4⋅33−6⋅32+4⋅3−1⋮=4n3−6n2+4n−1
上面各式左右两边分别相加，得到
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J=1^3+2^3+\cdots+n^3
J=13+23+⋯+n3）
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\begin{aligned} n^4-1^4 &=4J-4-6I+6+\frac{4n(n+1)}2-4-(n-1)\\ &=4J-6I-1+2n^2+n\\ &=4J-2n^3-n^2-1 \end{aligned}
n4−14=4J−4−6I+6+24n(n+1)−4−(n−1)=4J−6I−1+2n2+n=4J−2n3−n2−1
所以
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J=\frac{n^4+2n^3+n^2}4=\left[\frac{n(n+1)}{2}\right]^2.
J=4n4+2n3+n2=[2n(n+1)]2.
