Tính A :
A= (1-1/1+2)*(1-1/1+2+3)*(1-1/1+2+3+4)+...+(1-1/1+2+3+4+...+2016)
LA
Những câu hỏi liên quan
tính a: [1-1/2*2]*[1-1/3*3]*[1-1/4*4]*[1-1/5*5]*......*[1-1/2015*2015]*[1-1/2016*2016]
Cho
A = 1/2 + 1/3 + 1/4 + ... + 1/2017
B = 1/2016 + 2/2015 +3/2014+ ...+ 2015/2 + 2016/1
Tính B : A
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
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Câu 1
a) Chứng tỏ rằng 1/3 - 1/3^2 + 1/3^3 - 1/3^4 + 1/3^5 - 1/3^6 < 1/4
b) Cho A= 2015^2016 + 2016^2015 x 2015 và B= 1 + 2^2 + 3^2 + ......+2016^2. Tính AB có chia hết cho 5 không? Vì sao?
A = 1/2 + 1/3 +1/4 +.....+1/2016 + 1/2017 B = 2016/1 + 2015/2 + ......+ 2/2015 + 1/2016 . Tính B/A
\(\frac{B}{A}=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\frac{2017}{1}+\frac{2017}{2}+...+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\div\frac{1}{2017}=4068289\)
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Tính nhanh:
A= (1-1/2).(1-1/3).(1-1/4)...(1-1/2016)
B= (1/2-1).(1/3-1).(1/4-1)...(1/2016-1)
Ta có:
A=(1-1/2).(1-1/3).(1-1/4)...(1-1/2016)
=1/2.2/3.3/4...2015/2016
=1.2.3.4....2014.2015/2.3.4....2015.2016(Giống nhau bạn gạch đi)
=1/2016
Vậy A=1/2016
K cho mình nha :D
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TÍNH GIÁ TRỊ BIỂU THỨC A=2016+(2016/1+2)+(2016/1+2+3)+....+(2016/1+2+3+4+...+2016)
A = (1 - 1/(2*2)) * (1 - 1/(3*3)) * (1 - 1/(4*4)) ….* (1 - 1/(2016*2016)) .So sánh A với 1/2
so sánh A với 1/2 biết A = ( 1-1/2*2)* (1-1/3*3) * ( 1-1/4*4)* ...* ( 1-1/2016*2016)
A=\(\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{2016^2-1}{2016^2}\)
A=\(\frac{\left(2+1\right)\left(2-1\right)}{2^2}.\frac{\left(3+1\right)\left(3-1\right)}{3^2}......\frac{\left(2016+1\right)\left(2016-1\right)}{2016^2}\)
A=\(\frac{3.4......2017}{2.3....2016}.\frac{1.2...2015}{2.3...2016}\)
A=\(\frac{2017}{2}.\frac{1}{2016}\)
A=\(\frac{2017}{2.2106}>\frac{1}{2}\)
Vậy A\(>\frac{1}{2}\)
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Cho A = 1/2 + 1/3 + 1/4 + ... + 1/2017 B = 1/2015 + 2/2014 +3/2013 + ...+ 2015/2 + 2016/1 Tính B : A
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
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Tính :
\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)
\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)
\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{2016.2017:2}}\)
\(A=\frac{4032}{1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{3.4}+...+\frac{2}{2016.2017}}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2016}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{2015}{2017}\right)}\)
\(\Rightarrow A=2017\)
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\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)
\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{2016.2017:2}}\)
\(A=\frac{4032}{\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{2}{2016.2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2016}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+\frac{2015}{2017}}\)
\(A=2017\)
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