定義：求和記號∑ \sum ∑ ∑ k = 1 n a k = a 1 + a 2 + ⋯ + a n \sum_{k=1}^{n}{a_k}=a_1+a_2+\cdots+a_n k = 1 ∑ n a k = a 1 + a 2 + ⋯ + a n k k k ∑ k = m n c a k = c ∑ k = m n a k \sum_{k=m}^{n}{ca_k}=c\sum_{k=m}^{n}{a_k} k = m ∑ n c a k = c k = m ∑ n a k ∑ k = m n a k + b k = ∑ k = m n a k + ∑ k = m n b k \sum_{k=m}^{n}{a_k + b_k}=\sum_{k=m}^{n}{a_k}+\sum_{k=m}^{n}{b_k} k = m ∑ n a k + b k = k = m ∑ n a k + k = m ∑ n b k ∑ i = m n ∑ j = p q a i b j = ( ∑ i = m n a i ) ( ∑ j = p q b j ) = ∑ j = p q ∑ i = m n a i b j \sum_{i=m}^{n}\sum_{j=p}^{q}{a_i}{b_j}=\left(\sum_{i=m}^{n}{a_i}\right)\left(\sum_{j=p}^{q}{b_j}\right)=\sum_{j=p}^{q}\sum_{i=m}^{n}{a_ib_j} i = m ∑ n j = p ∑ q a i b j = ( i = m ∑ n a i ) ( j = p ∑ q b j ) = j = p ∑ q i = m ∑ n a i b j
等比數列a , a r , ⋯  , a r k , ⋯ a,ar,\cdots,ar^k,\cdots a , a r , ⋯ , a r k , ⋯ n + 1 n+1 n + 1 S = ∑ j = 0 n a r j = { ( n + 1 ) a r = 1 a r n + 1 − a r − 1 r ≠ 1 S=\sum_{j=0}^{n}{ar^j}=\left\{\begin{aligned}\left(n+1\right)a\quad\quad r=1\\\frac{ar^{n+1}-a}{r-1}\quad\quad r\neq1\end{aligned}\right. S = j = 0 ∑ n a r j = ⎩ ⎪ ⎨ ⎪ ⎧ ( n + 1 ) a r = 1 r − 1 a r n + 1 − a r ̸ = 1
定義：求積記號∏ \prod ∏ ∏ k = 1 n a k = a 1 a 2 ⋯ a n \prod_{k=1}^{n}{a_k}=a_1a_2\cdots a_n k = 1 ∏ n a k = a 1 a 2 ⋯ a n n ! : = ∏ j = 1 n j n ∈ N + n!:=\prod_{j=1}^{n}{j}\quad\quad n\in\N^+ n ! : = j = 1 ∏ n j n ∈ N +
約定：0 ! = 1 0!=1 0 ! = 1
