\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{n}-\frac{1}{n+3}\)
\(=1-\frac{1}{n+3}<1\)
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\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\)
a) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\frac{125}{376}\left(x€N\cdot\right)\)
b) \(\frac{3}{4}x-14\frac{2}{3}:\left(\frac{11}{15}+\frac{1111}{3535}+\frac{111111}{636363}\right)=12\)
Tìm x, biết:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
So sánh A và B biết
a,A=6\(\left(x+\frac{1}{3}\right)^2\)
,B=-8-(3,75-x)2
b,A=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
B=\(\left(\frac{1}{2}\right)^4\)
Chứng minh rằng:
a) \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\left(n,a\in Nsao\right)\)
4 tháng 9 2020 lúc 16:39
Chứng minh: \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{\left(n+1\right)}< 1\)với n\(\in\)N*
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\left(n\in N\right)\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+.......+\frac{1}{20}\left(1+2+3+4....+20\right)\)
Chứng minh:
a, \(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\frac{1}{n\left(n+2\right)}\right)< 2\)
b, \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< \frac{5}{4}\)
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.......+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\)(x ϵ N*)
Tìm x :