版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/u014157109/article/details/89437547
自从约瑟夫·傅立叶于1820年引入求和符号∑（大写的希腊字母sigma）以来，求和∑以及双重求和∑∑在数学公式推导，命题证明中被经常使用，掌握它的定义和性质对于提高我们的数学能力是必不可少的。有限项 的求和。
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\sum_{i=1}^{n}( a_{i}+b_{i})=\sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} b_{i}
i = 1 ∑ n ( a i + b i ) = i = 1 ∑ n a i + i = 1 ∑ n b i
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\sum_{i=1}^{n} r a_{i}=r \sum_{i=1}^{n} a_{i} \quad( r为任意常数)
i = 1 ∑ n r a i = r i = 1 ∑ n a i ( r 为 任 意 常 数 )
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\sum_{i=1}^{10} g(k, l) f(i, j)=g(k, l) \sum_{i=1}^{10} f(i, j)
i = 1 ∑ 1 0 g ( k , l ) f ( i , j ) = g ( k , l ) i = 1 ∑ 1 0 f ( i , j )
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\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{m} a_{i}+\sum_{i=m+1}^{n} a_{i}
i = 1 ∑ n a i = i = 1 ∑ m a i + i = m + 1 ∑ n a i
有一个n行m列的数表：
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\begin{array}{l}{a_{11}, a_{12}, a_{13}, \cdots, a_{1 m}} \\ {a_{21}, a_{22}, a_{23}, \cdots, a_{2 m}} \\ {a_{31}, a_{32}, a_{33}, \cdots, a_{2 m}} \\ {\cdots} \\ {a_{n 1}, a_{n 2}, a_{n 3}, \cdots, a_{n m}}\end{array}
a 1 1 , a 1 2 , a 1 3 , ⋯ , a 1 m a 2 1 , a 2 2 , a 2 3 , ⋯ , a 2 m a 3 1 , a 3 2 , a 3 3 , ⋯ , a 2 m ⋯ a n 1 , a n 2 , a n 3 , ⋯ , a n m
数表里的每个元素都由两个相互独立的数i，j决定，即每一项都是i，j的二元函数，一般项为aij ，i = 1,2…n; j = 1,2…m
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\sum_{j=1}^{m}(\sum_{i=1}^{n} a_{ij})
∑ j = 1 m ( ∑ i = 1 n a i j )
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\sum_{i=1}^{n}(\sum_{j=1}^{m} a_{ij})
∑ i = 1 n ( ∑ j = 1 m a i j )
第i行的元素的和记为Ri：
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R_{i} = \sum_{j=1}^{m} a_{ij} = a_{i1} + a_{i2} + ... + a_{im}
R i = j = 1 ∑ m a i j = a i 1 + a i 2 + . . . + a i m
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S = \sum_{i=1}^{n}R_{i} = \sum_{i=1}^{n}(\sum_{j=1}^{m} a_{ij})
S = i = 1 ∑ n R i = i = 1 ∑ n ( j = 1 ∑ m a i j )
第j列的元素的和记为Cj：
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C_{j} = \sum_{i=1}^{n} a_{ij} = a_{1j} + a_{2j} + ... + a_{nj}
C j = i = 1 ∑ n a i j = a 1 j + a 2 j + . . . + a n j
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S = \sum_{j=1}^{m}C_{j} = \sum_{j=1}^{m}(\sum_{i=1}^{n} a_{ij})
S = j = 1 ∑ m C j = j = 1 ∑ m ( i = 1 ∑ n a i j )
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\sum_{i=1}^{n}(\sum_{j=1}^{m} a_{ij}) = \sum_{j=1}^{m}(\sum_{i=1}^{n} a_{ij})
i = 1 ∑ n ( j = 1 ∑ m a i j ) = j = 1 ∑ m ( i = 1 ∑ n a i j )
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\sum_{1<=i<=n,1<=j<=m}a_{ij}
∑ 1 < = i < = n , 1 < = j < = m a i j
但要注意，但求和项数变为无穷或者（一个或两个）和号变为积分号时，往往要求级数收敛或者函数可积，相应的交换和号的结论才能成立。
Example 1
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\sum_{i=1}^{4} \sum_{j=1}^{i} f(i, j)
∑ i = 1 4 ∑ j = 1 i f ( i , j )
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\sum_{j=1}^{i} \sum_{i=1}^{4} f(i, j)
∑ j = 1 i ∑ i = 1 4 f ( i , j )
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\sum_{j=1}^{4} \sum_{i=1}^{j} f(i, j)
∑ j = 1 4 ∑ i = 1 j f ( i , j )
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\sum_{j=1}^{4} \sum_{i=j}^{4} f(i, j)
∑ j = 1 4 ∑ i = j 4 f ( i , j )
因为[1<= i <= 4][1<= j<= i] = [1<= j <= i <= 4] = [1<= j<= 4][ j <= i <= 4]
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\sum_{i=1}^{4} \sum_{j=1}^{i} f(i, j)
∑ i = 1 4 ∑ j = 1 i f ( i , j ) 下三角 矩阵的形式（按行求和），然后将按行求和切换为按列求和，也会得到C答案。
Example 2
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\sum_{k=1}^{n} \sum_{i=1}^{k} \frac{ ia_{ij}}{k(k + 1)}
∑ k = 1 n ∑ i = 1 k k ( k + 1 ) i a i j
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\sum_{i=1}^{n} \sum_{k=i}^{n} \frac{ ia_{ij}}{k(k + 1)}
∑ i = 1 n ∑ k = i n k ( k + 1 ) i a i j
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\sum_{k=1}^{n} \sum_{i=1}^{k} \frac{ ia_{ij}}{k(k + 1)}
∑ k = 1 n ∑ i = 1 k k ( k + 1 ) i a i j 上三角 矩阵的形式（按列求和），然后将按列求和切换为按行求和。
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Example 3
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\sum_{k=1}^{n}(k( \sum_{i=1}^{k}\frac{ i^2}{a_{i}}) (\frac{2}{k(k + 1)})^2)
∑ k = 1 n ( k ( ∑ i = 1 k a i i 2 ) ( k ( k + 1 ) 2 ) 2 )
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4\sum_{k=1}^{n} \sum_{i=1}^{k}\frac{ i^2}{a_{i}} \frac{k}{k(k + 1)^2}
4 ∑ k = 1 n ∑ i = 1 k a i i 2 k ( k + 1 ) 2 k
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4\sum_{i=1}^{n}\frac{ i^2}{a_{i}} \sum_{k=i}^{n} \frac{k}{k(k + 1)^2}
4 ∑ i = 1 n a i i 2 ∑ k = i n k ( k + 1 ) 2 k
注意 ，容易出错的地方
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\left(\sum_{i=1}^{5} f(i)\right)^{2}=\left(\sum_{i=1}^{5} f(i)\right) *\left(\sum_{i=1}^{5} f(i)\right) ≠ \sum_{i=1}^{5} \sum_{i=1}^{5} f(i) f(i) = \sum_{i=1}^{5} \sum_{i=1}^{5} f^2(i)
( ∑ i = 1 5 f ( i ) ) 2 = ( ∑ i = 1 5 f ( i ) ) ∗ ( ∑ i = 1 5 f ( i ) ) ̸ = ∑ i = 1 5 ∑ i = 1 5 f ( i ) f ( i ) = ∑ i = 1 5 ∑ i = 1 5 f 2 ( i )
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\left(\sum_{i=1}^{5} f(i)\right)^{2}=\left(\sum_{i=1}^{5} f(i)\right) *\left(\sum_{i=1}^{5} f(i)\right) = \sum_{i=1}^{5} \sum_{j=1}^{5} f(i) f(j)
( ∑ i = 1 5 f ( i ) ) 2 = ( ∑ i = 1 5 f ( i ) ) ∗ ( ∑ i = 1 5 f ( i ) ) = ∑ i = 1 5 ∑ j = 1 5 f ( i ) f ( j )
–更详细内容可阅读《具体数学》第二章 和式
