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\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30
4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)
4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30
4A = 28.29.30.31 - 0.1.2.3
4A = 28.29.30.31
\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)
Theo cách tính trên ta dễ dàng tính được:
1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
C = 1.2.3+ 2.3.4 + 3.4.5 +...+n(n+1) ( n+2)
\(\Rightarrow4C=1.2.3\left(4-0\right)+2.3.4.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-\left(n-1\right)\right]\)
\(=1.2.3.4-0.1.2.3+2.3.4.5-...+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n+2\right)\) \(=n\left(n+1\right)\left(n+2\right)\left(n+3\right)-0.1.2.3\)
\(=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)
\(\Rightarrow C=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)
Đặt S = 1/1.2.3 - 1/2.3.4 - 1/3.4.5 - ...- 1/97.98.99
S x 2 = 2/1.2.3 - 2/2.3.4 - 2/3.4.5 - ...- 2/97.98.99
= (1/1.2 -1/2.3) - (1/2.3 - 1/3.4 ) - (1/3.4 - 1/4.5) - ...- (1/97.98 - 1/98.99)
= 1/1.2 - 1/2.3 - 1/2.3 + 1/3.4 - 1/3.4 + 1/4.5 - ....- 1/97.98 + 1/98.99
= 1/2 -1/3 + 1/98.99
= 1618/9072 => S = 1618/9072 : 2 = 809/9072
Ta có \(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
Áp dụng:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{10\cdot11\cdot12}\\ =\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{10\cdot11}-\dfrac{1}{11\cdot12}\\ =\dfrac{1}{2}-\dfrac{1}{11\cdot12}=\dfrac{1}{2}-\dfrac{1}{132}=\dfrac{65}{132}\)
sai rồi kìa
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3}\) mà