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Ta có :
\(S=2015+\frac{2015}{1+2}+\frac{2015}{1+2+3}+...+\frac{2015}{1+2+3+..+2016}\)
\(=2015.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}\right)\)
\(=2015.\left(1+\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(2016+1\right).2016}{2}}\right)\)
\(=2015.\left(\frac{2}{2}+\frac{2}{2.\left(2+1\right)}+\frac{2}{3.\left(3+1\right)}+...+\frac{2}{2016.\left(2016+1\right)}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2.\left(2+1\right)}+\frac{1}{3.\left(3+1\right)}+...+\frac{1}{2016.\left(2016+1\right)}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{2017}\right)\)
\(=2015.2.\left(1-\frac{1}{2017}\right)\)
\(=2015.2.\frac{2016}{2017}\)
=\(\frac{2015.2.2016}{2017}\)
=\(\frac{8124480}{2017}\)
Vậy \(S=\frac{8124480}{2017}\)
Thay a,b,c lần lượt vào biểu thức...
Tính được kết quả:
a) A= \(-\frac{7}{10}\)
b) B= \(-\frac{2}{7}\)
c) C= 0
A=-1/2*-2/3*-3/4*..*-2013/2014
A=-1*-2*-3*...*-2013/2*3*4*...*2014
A=-1/2014
ta có(-1)^2015=-1
B=-1/2015>-1/2014=A
nên A<B
A=\(\frac{2014}{2014^a}+\frac{2014}{2014^b}\)=B=\(\frac{2013}{2015^a}\)+\(\frac{2015}{2013^b}\)
Ta có: 2014/\(2014^a\)+2014/2014^b= 2013/2014^a + 1/2014^a +2015/2014^a - 1/2014^a
=(2013/2014^a + 2015/2014^b) + ( 1/2014^a + 1/2014^b)
= B + (1/2014^a + 1/2014^b)
*Nếu a=b thì A=B
*Nếu a>b thì (1/2014^a + 1/2014^b) >0
\(\Rightarrow\) A< B
*Nếu a<b thì (1/2014^a + 1/2014^b)>0
\(\Rightarrow\) A>B
B= \(\frac{1}{199}\) + \(\frac{2}{198}\) + ... + \(\frac{198}{2}\) + \(\frac{199}{1}\)
B= ( \(\frac{1}{199}\) + 1) + ( \(\frac{2}{198}\) +1) +...+ ( \(\frac{198}{2}\) +1) +1 ( Mình tách 199 ra thành 199 số hạng rồi cộng thêm vào mỗi phân số)
B= \(\frac{200}{199}\) + \(\frac{200}{198}\) + \(\frac{200}{197}\) +...+\(\frac{200}{2}\)
B= 200( \(\frac{1}{199}\) + \(\frac{1}{198}\) +...+ \(\frac{1}{2}\) )
B= 200 ( \(\frac{1}{2}\) + \(\frac{1}{3}\) +...+ \(\frac{1}{198}\) + \(\frac{1}{199}\) ) = 200 A
Ta thấy A=1A, B=200A Suy ra \(\frac{A}{B}\) = \(\frac{1}{200}\)
Không chép lại đề nhé
Ta có:
P=\(\frac{50-49}{49}+\frac{50-48}{48}+...+\frac{50-2}{2}+\frac{50-1}{1}\)
P=\(\frac{50}{49}-\frac{49}{49}+\frac{50}{48}-\frac{48}{48}+...+\frac{50}{2}-\frac{2}{2}+\frac{50}{1}-\frac{1}{1}\)
P=\(\left(\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\right)+\frac{50}{1}-\left(\frac{49}{49}+\frac{48}{48}+...+\frac{2}{2}+\frac{1}{1}\right)\)
P=\(50\cdot\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)+50-49\) (chỗ này gộp nha)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{48}+\frac{1}{49}\right)+1\)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)+\frac{50}{50}\)
P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)\)
=>P=50S
=>\(\frac{S}{P}=\frac{S}{50S}=\frac{1}{50}\)
Vừa nãy mình nói nhầm, Sorry.
\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)
\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)
a) \(\frac{1}{n}\) - \(\frac{1}{n+1}\) = \(\frac{n+1}{n\left(n+1\right)}\) - \(\frac{n}{n\left(n+1\right)}\) = \(\frac{1}{n\left(n+1\right)}\) = \(\frac{1}{n}\) . \(\frac{1}{n+1}\) =>đpcm
b) A= \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{8}\) - \(\frac{1}{9}\) +\(\frac{1}{9}\)
= \(\frac{1}{2}\) + \(\frac{1}{9}\)= \(\frac{11}{18}\)
2016