E=1/1.2.3.4+1/2.3.4.5+...…+1/7.8.9.10
Các bạn giúp mình nhé mình cảm ơn rất nhiều
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NT
2
10 tháng 8 2017
Ta xét:
\(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}=\dfrac{3}{1.2.3.4};\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}=\dfrac{3}{2.3.4.5};.....;\dfrac{1}{7.8.9}-\dfrac{1}{8.9.10}=\dfrac{3}{7.8.9.10}\)
Gọi biểu thức phải tính là A, ta có:
3A=\(\dfrac{1}{1.2.3}-\dfrac{1}{8.9.10}=\dfrac{714}{4320}\)
Vậy A=\(\dfrac{238}{1440}\)
bài này không 1\299 + 4\299+7\299 +...... + 298\299
VT
10 tháng 10 2019
a)\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{1}{n+1}.\left(\frac{1}{n}-\frac{1}{n+2}\right)\)=\(\frac{1}{2}.\frac{1}{n\left(n+1\right)}-\frac{1}{2}.\frac{1}{\left(n+1\right)\left(n+2\right)}\)= \(\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)
=> a = \(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)\)+\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}\right)\)+....+\(\frac{1}{2}\left(\frac{1}{2018}-\frac{1}{2019}\right)-\frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)=\(\frac{1}{2}\left(1-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)=\(\frac{1}{4}\left(1-\frac{1}{2019.1010}\right)\)=\(\frac{2019.1010-1}{2.2019.2020}\)
b) tương tự \(\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\)=\(\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)-\(\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)\)=\(\frac{1}{6}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{3}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)+\(\frac{1}{6}\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\)= M-P+N
Với n từ 1 đến 2017 thì
M= \(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{2}\right)+\frac{1}{6}\left(\frac{1}{2}-\frac{1}{3}\right)+...\)+\(\frac{1}{6}\left(\frac{1}{2017}-\frac{1}{2018}\right)\)=\(\frac{1}{6}\left(1-\frac{1}{2018}\right)=\frac{2017}{6.2018}\)
N= \(\frac{1}{6}\left(\frac{1}{3}-\frac{1}{4}\right)+\frac{1}{6}\left(\frac{1}{4}-\frac{1}{5}\right)+...+\)\(\frac{1}{6}\left(\frac{1}{2019}-\frac{1}{2020}\right)=\)\(\frac{1}{6}\left(\frac{1}{3}-\frac{1}{2020}\right)=\frac{2017}{6.3.2020}\)
P= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{3}-\frac{1}{4}\right)+...+\)\(\frac{1}{3}\left(\frac{1}{2018}-\frac{1}{2019}\right)\)= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{2019}\right)=\frac{2017}{3.2.2019}\)
M+N-P = \(\frac{2017}{6}\left(\frac{1}{2018}+\frac{1}{3.2020}-\frac{1}{2019}\right)\)=\(\frac{2017}{6}.\left(\frac{1}{2018.2019}+\frac{1}{3.2020}\right)\)
= \(\frac{2017\left(1010+1009.673\right)}{3.2018.2019.2020}\)
TB
0
NT
0
10 tháng 9 2015
=1/1.2-1/3.4+1/2.3-1/3.4+...+1/116.117-1/118.119
=1-1/2-1/3+1/4+1/2-1/3-1/3-1/4+...+1/116-1/117-1/118+1/119
=1+1/119=120/119(ko nhầm thì z)
23 tháng 10 2018
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 45 + 10 = 55
* Hok tốt !
T1
23 tháng 10 2018
=(1+9)+(2+8)+(3+7)+(4+6)+10+5
=10+10+10+10+10+5
=10.5+5
=50+5
=55
