Ta có: \(55+5\)1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 +.....+ 1/50^2 = 1/1^2 + 1/2^2 + (1/3^2 + 1/4^2 +....+ 1/50^2 )
< 1 + 1/4 + (1/2*3 + 1/3*4 +...+1/49*50) = 1 + 1/4 + (1/2 - 1/3 + 1/3 - 1/4+...+1/49 - 1/50 )
= 1,73 = 173/100 (dpcm)
Cmr:\(50< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{100-1}}< 100\)
CMR:
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2019\times2020}< 1\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{3}{4}\)
\(C=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}< 2\)
Cho:
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2^{100}-1}\)
CMR: A>50
CMR: 100-(\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\))=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
CMR:
a,\(100\left(1+\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+........+\frac{99}{100}\)
CMR:
a) \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{100^2}< 1\)
b)\(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{100^2}< 1\frac{3}{4}\)
c)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
CMR :
a , A = \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+.....+\frac{19}{9^2.10^2}< 1\)
b , B = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+......+\frac{100}{3^{100}}< \frac{3}{4}\)
c, C = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)< \frac{1}{2}\)
Cầu giúp đỡ ~~~
CMR:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{3}{4}\)
