Tìm n thuộc n :
1/3.4 + 1/4.5 + 1/5.6 + ... + 1/n( n+1 ) = 3/10
VP
Những câu hỏi liên quan
Tìm n thuộc N :
1/3.4 + 1/4.5 + 1/5.6 + .... + 1/n( n+1 ) = 3/10
\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + .....+\(\dfrac{1}{n.(n+1)}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\) +......+ \(\dfrac{1}{n}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{3}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{n+1}\) = \(\dfrac{1}{3}-\dfrac{3}{10}\)
\(\dfrac{1}{n+1}\) = \(\dfrac{1}{30}\)
n + 1 = 30
n = 30 - 1
n = 29
Kết luận n = 29 là giá trị thỏa mãn yêu cầu đề bài.
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tìm n thuộc N biết:
\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+....+\(\dfrac{1}{n\left(n+1\right)}\)=\(\dfrac{3}{10}\)
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{n+1}=\dfrac{1}{30}\)
\(\Rightarrow n+1=30\)
\(\Rightarrow n=29\)
Vậy n = 29.
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1/3.4 +1/4.5 + 1/5.6 + .. + 1/n.(n+1) = 3/10
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)
Ta có: \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{3}{10}\)
\(\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{3}{10}\)
\(\dfrac{1}{x+1}=\dfrac{1}{30}\)
\(\Rightarrow x+1=30\)
\(x=30-1\)
\(x=29\)
Vậy ...
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Cho M=1/2.3/4.5/6...99.100 va N=2/3.4/5.6/7.... 100/101 va N=2/3.4/5.6/7..100/101
a) So sanh M va N b) Tinh M.N c) So sanh M va 1/10
cho M= 1/2.3/4.5/6.....99/100 và N= 2/3.4/5.6/7.....100/101
chứng minh m<n
tìm tích m.n
chưng minh m<1/10
tính giá trị biểu thức
A =\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
B = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{n.\left(n+1\right)}\)(n\(\in\)Z, n\(\ne\)0; n\(\ne\)-1)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}=\frac{5}{6}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)
ui cí này e chưa học
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}\)
\(=\frac{5}{6}\)
Xem thêm câu trả lời
c) 1/3.4 + 1/4,5 + 1/5.6 + .... + 1/x.(x + 1) = 3/10 và x thuộc N
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{30}\)
\(\Rightarrow x=30-1\)
\(\Rightarrow x=29\)
vậy: \(x=29\)
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tìm x, biết
1/3.4+1/4.5+1/5.6+1/6.7+...+1/x.(x+1)=3/10
(1/3-1/4+1/4-1/5+1/5-.......+1/x.(x+1)=3/10
1/3-1/x+1=3/10
tự làm...
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