Chứng minh \(\frac{3}{1^2x2^2}+\frac{5}{2^2x3^2}+\frac{7}{3^2x4^2}+...+\frac{2013}{1006^2x1007^2}\) < 1
\(A=\frac{3}{1^2x2^2}+\frac{5}{2^2x3^2}+\frac{7}{3^2x4^2}+............+\frac{19}{9^2x10^2}\)
\(A=\frac{3}{1^2x2^2}+\frac{5}{2^2x3^2}+\frac{7}{3^2x4^2}+......+\frac{39}{19^2x20^2}\)
Tính A
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{39}{19^2.20^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{19^2}-\frac{1}{20^2}\)
\(=1-\frac{1}{20^2}\)
\(=\frac{399}{400}\)
Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+.....+\frac{2013}{2006^2.2007^2}<1\)
tính:
a) \(\frac{1^2}{1x2}+\frac{2^2}{2x3}+\frac{3^2}{3x4}+...+\frac{100^2}{100x101}\)
b) \(\frac{2^2}{1x3}+\frac{3^2}{2x4}+\frac{4^2}{3x5}+...+\frac{59^2}{58x60}\)
a) Đặt \(A=\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}\)
\(\Rightarrow A=\left(1^2+2^2+..........+100^2\right)\)\(.\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+.....+100^2\right).\left(\frac{100}{101}\right)\)(a)
Đặt \(M=\left(1^2+2^2+........+100^2\right)\)
\(\Rightarrow M=1.1+2.2+.....+100.100\)
\(\Rightarrow M=1.\left(2-1\right)+2.\left(3-1\right)+....+100.\left(101-1\right)\)
\(\Rightarrow M=\left(1.2-1\right)+\left(2.3-2\right)+.....+\left(100.101-100\right)\)
\(\Rightarrow M=\left(1.2+2.3+.....+100.101\right)-\left(1+2+......+100\right)\)
\(\Rightarrow M=\left(1.2+2.3+......+100.101\right)-5050\)(1)
Đặt \(N=1.2+2.3+....+100.101\)
\(\Rightarrow3.N=1.2.3+2.3.3+......+100.101.3\)
\(\Rightarrow3N=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+100.101.\left(102-99\right)\)
\(\Rightarrow3N=\left(1.2.3-0\right)+\left(1.2.3-2.3.4\right)+.......+\left(100.101.102-100.101.99\right)\)
\(\Rightarrow3N=100.101.102-0\)
\(\Rightarrow N=343400\)
Thay N = 343400 vào 1) ta được:
M = 343400 - 5050
=> M = 338350
Thay M = 338350 Vào (a) ta được:
A = 338350 . \(\frac{100}{101}\)
=> \(A=\frac{33835000}{101}\)
Vậy \(\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}=\frac{33835000}{101}=335000\)
b) Đặt \(B=\frac{2^2}{1.3}+\frac{3^2}{2.4}+..........+\frac{59^2}{58.60}\)
\(\Rightarrow B=\left(2^2+3^2+........+59^2\right).\left(\frac{1}{1.3}+\frac{1}{2.4}+.....+\frac{1}{58.60}\right)\)
Đặt \(G=2^2+3^2+.........+59^2\)VÀ \(H=\frac{1}{1.3}+\frac{1}{2.4}+.........+\frac{1}{58.60}\)
\(\Rightarrow G=2.2+3.3+.......+59.59\) VÀ \(2.H=\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{58.60}\)
Rồi bạn làm như ở phần a) ý
chứng minh\(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2013}+\frac{1}{2014}< \frac{2}{5}\)
chứng minh \(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2012}-\frac{1}{2013}+\frac{1}{2014}< \frac{2}{5}\)
Chứng minh: \(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+..+\frac{n}{2^n}+...+\frac{2013}{2^{2013}}+\frac{2014}{2^{2014}}<2\)
Giải các phương trình và bất phương trình sau
a) \(\frac{x-1}{2015}+\frac{x}{2014}+\frac{2}{1006}=\)\(\frac{x-3}{2013}+\frac{x}{2012}+\frac{1}{1007}\)
b) \(\frac{4}{1+y+y^2}+\frac{1}{1-y}\le\frac{2y^2-5}{y^3-1}\)
chứng minh: \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2013^2}< \frac{1}{4}\)\(\frac{1}{4}\)