Tính giá trị biểu thức sau:
A = \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right).\)
K
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17 tháng 5 2017
=1/2.(1+1/1.3).(1+1/2.4).(1+1/3.5)...(1+1/2014.2016)
=1/2.(1+1/1-1/3).(1+1/3-1/5)...(1+1/2014-1/2016)
=1/2.1+(1/1-1/2016)
=1/2.2015/2016
=2015/4032
BM
8 tháng 4 2016
Ta có :
\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}............\frac{2015^2}{2014.2016}\)= \(\frac{2.2}{1.3}.\frac{3.3}{2.4}...........\frac{2015.2015}{2014.2016}=\frac{2.2015}{2016}=\frac{2015}{1008}\)
k cho mình nha
19 tháng 3 2017
Ta có công thức :
\(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Áp dụng vào bài toán ta được :
\(C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}..........\frac{2015^2}{2014.2016}\)
\(=\frac{\left(2.3.4....2015\right)\left(2.3.4....2015\right)}{\left(1.2.3...2014\right)\left(3.4.5.....2016\right)}\)
\(=\frac{2015.2}{2016}=\frac{2015}{1008}\)
19 tháng 3 2017
=1(1/1*3*(1/2*4)*...*(1+1/2014*2016)
=1/2(2+2/1*3)+(2+2/2*4)*...(2+2/2014*2016)
=1/2(2+1/1-1/3)...(2+1/2014-1/2016)
=1/2*(1/1-1/2016)
=3023/4032
S
10 tháng 4 2018
\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2013\cdot2015}\right)\)
\(=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{4056196}{2013\cdot2015}\)
\(=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}\)
\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}\)
\(=\frac{2014\cdot2}{1\cdot2015}\)
\(=\frac{4028}{2015}\)
9 tháng 4 2017
2A=\(\left(1+\frac{1}{3}\right)\)\(\left(1+\frac{1}{8}\right)\)\(\left(1+\frac{1}{15}\right)\)\(.......\)\(\left(1+\frac{1}{4064255}\right)\)
2A = \(\frac{4}{3}\)\(.\)\(\frac{9}{8}\)\(.\)\(\frac{16}{15}\)\(......\)\(\frac{4064256}{4064255}\)
2A = \(\frac{2.2}{1.3}\)\(.\)\(\frac{3.3}{2.4}\)\(.\)\(\frac{4.4}{3.5}\)\(......\)\(\frac{2016.2016}{2015.2017}\)
2A = \(\frac{2.3.4....2016}{1.2.3.....2015}\)\(.\)\(\frac{2.3.4....2016}{3.4.5....2017}\)
2A = \(\frac{2016}{1}\)\(.\)\(\frac{2}{2017}\)
2A = \(\frac{4032}{2017}\)
A = \(\frac{4032}{2017}\)\(:2\)
A = \(\frac{2016}{2017}\)
29 tháng 3 2016
Ta có :
1 + 1/(1.3) = (1.3 + 1)/(1.3) = [(2 - 1)(2 + 1) + 1]/(1.3) = 2^2 / (1.3)
1 + 1/(2.4) = (2.4 + 1)/(2.4) = [(3 - 1)(3 + 1) + 1]/(2.4) = 3^2 / (2.4)
1 + 1/(3.5) = (3.5 + 1)/(3.5) = [(4 - 1)(4 + 1) + 1]/(3.5) = 4^2 / (3.5)
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1+1/(2014.2016)=(2014.2016+1)/(2014.2016)=[(2015-1)(2015+1)+1]/(2014.2016)=2015^2/(2014.2016)
Nhân các bất đẳng thức trên vé theo vế ta được:
------->S= (2^2) (3^2) (4^2) ...(2015^2)/[(1.3) (2.4) (3.5) ...(2013.2015)(2014.2016)]
S= (2^2) (3^2) (4^2) ... (2015^2) / [1.2.(3^2)(4^2)(5^2) ... (2013^2)(2014^2).2015.2016]
S = (2.2015) / 2016
Cái này là đáp số rồi...Đúng 100% đó bạn k giúp mình nhé :P
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)
\(A=\frac{2.3.4...2015}{1.2.3...2014}.\frac{2.3.4...2015}{3.4.5...2016}\)
\(A=2015.\frac{1}{1008}\)
\(A=\frac{2015}{1008}\)