ตัวอย่างต่อไปนี้เป็นเอกลักษณ์ที่เกี่ยวกับผลรวมที่สำคัญ
∑ n = s t C ⋅ f ( n ) = C ⋅ ∑ n = s t f ( n ) {\displaystyle \sum _{n=s}^{t}C\cdot f(n)=C\cdot \sum _{n=s}^{t}f(n)} เมื่อ C เป็น
ค่าคงตัว (ดูเพิ่มที่
การคูณสเกลาร์) ∑ i = s n f ( C ) = ( n − s + 1 ) f ( C ) {\displaystyle \sum _{i=s}^{n}f(C)=(n-s+1)f(C)} เมื่อ C เป็นค่าคงตัว ∑ n = s t f ( n ) + ∑ n = s t g ( n ) = ∑ n = s t [ f ( n ) + g ( n ) ] {\displaystyle \sum _{n=s}^{t}f(n)+\sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left[f(n)+g(n)\right]} ∑ n = s t f ( n ) = ∑ n = s + p t + p f ( n − p ) {\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=s+p}^{t+p}f(n-p)} ∑ n = s j f ( n ) + ∑ n = j + 1 t f ( n ) = ∑ n = s t f ( n ) {\displaystyle \sum _{n=s}^{j}f(n)+\sum _{n=j+1}^{t}f(n)=\sum _{n=s}^{t}f(n)} ∑ i = m n x = ( n − m + 1 ) x {\displaystyle \sum _{i=m}^{n}x=(n-m+1)x} ∑ i = 1 n x = n x {\displaystyle \sum _{i=1}^{n}x=nx} เป็นนิยามของการคูณ เมื่อ n เป็นจำนวนเต็มซึ่งเป็นตัวคูณของ x ∑ i = m n i = ( n − m + 1 ) ( n + m ) 2 {\displaystyle \sum _{i=m}^{n}i={\frac {(n-m+1)(n+m)}{2}}} (ดูเพิ่มที่
อนุกรมเลขคณิต) ∑ i = 0 n i = ∑ i = 1 n i = ( n ) ( n + 1 ) 2 {\displaystyle \sum _{i=0}^{n}i=\sum _{i=1}^{n}i={\frac {(n)(n+1)}{2}}} (กรณีพิเศษของอนุกรมเลขคณิต) ∑ i = 1 n i 2 = n ( n + 1 ) ( 2 n + 1 ) 6 = n 3 3 + n 2 2 + n 6 {\displaystyle \sum _{i=1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}} ∑ i = 1 n i 3 = ( n ( n + 1 ) 2 ) 2 = n 4 4 + n 3 2 + n 2 4 = [ ∑ i = 1 n i ] 2 {\displaystyle \sum _{i=1}^{n}i^{3}=\left({\frac {n(n+1)}{2}}\right)^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}=\left[\sum _{i=1}^{n}i\right]^{2}} ∑ i = 1 n i 4 = n ( n + 1 ) ( 2 n + 1 ) ( 3 n 2 + 3 n − 1 ) 30 = n 5 5 + n 4 2 + n 3 3 − n 30 {\displaystyle \sum _{i=1}^{n}i^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}={\frac {n^{5}}{5}}+{\frac {n^{4}}{2}}+{\frac {n^{3}}{3}}-{\frac {n}{30}}} ∑ i = 0 n i p = ( n + 1 ) p + 1 p + 1 + ∑ k = 1 p B k p − k + 1 ( p k ) ( n + 1 ) p − k + 1 {\displaystyle \sum _{i=0}^{n}i^{p}={\frac {(n+1)^{p+1}}{p+1}}+\sum _{k=1}^{p}{\frac {B_{k}}{p-k+1}}{p \choose k}(n+1)^{p-k+1}} เมื่อ B k {\displaystyle B_{k}} เป็น
จำนวนแบร์นูลลีตัวที่ k
∑ i = m n x i = x n + 1 − x m x − 1 {\displaystyle \sum _{i=m}^{n}x^{i}={\frac {x^{n+1}-x^{m}}{x-1}}} (ดูเพิ่มที่
อนุกรมเรขาคณิต) ∑ i = 0 n x i = x n + 1 − 1 x − 1 {\displaystyle \sum _{i=0}^{n}x^{i}={\frac {x^{n+1}-1}{x-1}}} (กรณีพิเศษของสูตรก่อนหน้านี้ เมื่อ m = 0) ∑ i = 0 n i 2 i = 2 + 2 n + 1 ( n − 1 ) {\displaystyle \sum _{i=0}^{n}i2^{i}=2+2^{n+1}(n-1)} ∑ i = 0 n i 2 i = 2 n + 1 − n − 2 2 n {\displaystyle \sum _{i=0}^{n}{\frac {i}{2^{i}}}={\frac {2^{n+1}-n-2}{2^{n}}}} ∑ i = 0 n i x i = x ( 1 − x ) 2 ( x n ( n ( x − 1 ) − 1 ) + 1 ) {\displaystyle \sum _{i=0}^{n}ix^{i}={\frac {x}{(1-x)^{2}}}(x^{n}(n(x-1)-1)+1)} ∑ i = 0 n i 2 x i = x ( 1 − x ) 3 ( 1 + x − ( n + 1 ) 2 x n + ( 2 n 2 + 2 n − 1 ) x n + 1 − n 2 x n + 2 ) {\displaystyle \sum _{i=0}^{n}i^{2}x^{i}={\frac {x}{(1-x)^{3}}}(1+x-(n+1)^{2}x^{n}+(2n^{2}+2n-1)x^{n+1}-n^{2}x^{n+2})}
∑ i = 0 n ( n i ) = 2 n {\displaystyle \sum _{i=0}^{n}{n \choose i}=2^{n}} (ดูเพิ่มที่
สัมประสิทธิ์ทวินาม) ∑ i = 0 n − 1 ( i k ) = ( n k + 1 ) {\displaystyle \sum _{i=0}^{n-1}{i \choose k}={n \choose k+1}} ( ∑ i a i ) ( ∑ j b j ) = ∑ i ∑ j a i b j {\displaystyle \left(\sum _{i}a_{i}\right)\left(\sum _{j}b_{j}\right)=\sum _{i}\sum _{j}a_{i}b_{j}}
( ∑ i a i ) 2 = 2 ∑ i ∑ j < i a i a j + ∑ i a i 2 {\displaystyle \left(\sum _{i}a_{i}\right)^{2}=2\sum _{i}\sum _{j<i}a_{i}a_{j}+\sum _{i}a_{i}^{2}} ∑ n = a b f ( n ) = ∑ n = b a f ( n ) {\displaystyle \sum _{n=a}^{b}f(n)=\sum _{n=b}^{a}f(n)} ∑ n = s t f ( n ) = ∑ n = − t − s f ( − n ) {\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=-t}^{-s}f(-n)} ∑ n = 0 t f ( 2 n ) + ∑ n = 0 t f ( 2 n + 1 ) = ∑ n = 0 2 t + 1 f ( n ) {\displaystyle \sum _{n=0}^{t}f(2n)+\sum _{n=0}^{t}f(2n+1)=\sum _{n=0}^{2t+1}f(n)} ∑ n = 0 t ∑ i = 0 z − 1 f ( z ⋅ n + i ) = ∑ n = 0 z ⋅ t + z − 1 f ( n ) {\displaystyle \sum _{n=0}^{t}\sum _{i=0}^{z-1}f(z\cdot n+i)=\sum _{n=0}^{z\cdot t+z-1}f(n)} b ^ [ ∑ n = s t f ( n ) ] = ∏ n = s t b ^ f ( n ) {\displaystyle {\widehat {b}}^{\left[\sum _{n=s}^{t}f(n)\right]}=\prod _{n=s}^{t}{\widehat {b}}^{f(n)}} (ดูเพิ่มที่
ผลคูณของอนุกรม) ∑ n = s t ln f ( n ) = ln ∏ n = s t f ( n ) {\displaystyle \sum _{n=s}^{t}\ln f(n)=\ln \prod _{n=s}^{t}f(n)} lim t → ∞ ∑ n = a t f ( n ) = ∑ n = a ∞ f ( n ) {\displaystyle \lim _{t\rightarrow \infty }\sum _{n=a}^{t}f(n)=\sum _{n=a}^{\infty }f(n)} (ดูเพิ่มที่
ลิมิตอนันต์) ( a + b ) n = ∑ i = 0 n ( n i ) a ( n − i ) b i {\displaystyle (a+b)^{n}=\sum _{i=0}^{n}{n \choose i}a^{(n-i)}b^{i}} สำหรับ
การกระจายทวินาม ∑ n = b + 1 ∞ b n 2 − b 2 = ∑ n = 1 2 b 1 2 n {\displaystyle \sum _{n=b+1}^{\infty }{\frac {b}{n^{2}-b^{2}}}=\sum _{n=1}^{2b}{\frac {1}{2n}}} ( ∑ i = 1 n f i ( x ) ) ′ = ∑ i = 1 n f i ′ ( x ) {\displaystyle \left(\sum _{i=1}^{n}f_{i}(x)\right)^{\prime }=\sum _{i=1}^{n}f_{i}^{\prime }(x)} lim n → ∞ ∑ i = 0 n f ( a + b − a n i ) ⋅ b − a n = ∫ a b f ( x ) d x {\displaystyle \lim _{n\to \infty }\sum _{i=0}^{n}f\left(a+{\frac {b-a}{n}}i\right)\cdot {\frac {b-a}{n}}=\int _{a}^{b}f(x)dx}
