Các câu hỏi tương tự
Chứng minh rằng: \(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}
Bài 5 :
a) Tính giá trị của biểu thức :
\(A=\frac{\left(81,624:4\frac{4}{3}-4.505\right)^2+125\frac{3}{4}}{\left\{\left[\left(\frac{11}{25}\right)^2:0,88+3,53\right]^2-\left(2,75\right)^2\right\}:\frac{13}{25}}\)
b) Chứng minh rằng tổng :
\(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{4n-2}}-\frac{1}{2^n}+...+\frac{1}{2^{2002}-}-\frac{1}{2^{2004}}< 0,2\)
Bài 1 :
1 . Tính :
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
2 . Biết : 13 + 23 + ... + 103 = 3025
Tính : S = 23 + 43 + 63 + .... + 203
Chứng tỏ rằng:
1-\(\frac{15}{16}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{4n^2}< \frac{1}{4}\)
Chứng minh rằng:
a,\(\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}=\frac{5n}{3.\left(4n+3\right)}\)
b,\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}< \frac{1}{4}\)
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{4n^2}< \frac{1}{4}\)( với N \(\varepsilon\)N*)
Chứng minh rằng: \(A=\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{4n-2}}-\frac{1}{3^{4n}}+...+\frac{1}{3^{98}}-\frac{1}{3^{100}}
Chứng minh rằng:
\(A=\frac{1}{3^2}+\frac{1}{3^4}+......+\frac{1}{3^{4n-2}}+\frac{1}{3^{4n}}+...+\frac{1}{3^{98}}-\frac{1}{3^{100}}<0,1\)
Chứng minh rằng:
A=\(\frac{1}{3^2}+\frac{1}{3^4}+.......+\frac{1}{3^{4n-2}}+\frac{1}{3^{4n}}+....+\frac{1}{3^{98}}-\frac{1}{3^{100}}\)< 0,1