ta có: \(1=\frac{1}{1^2};\frac{1}{4}=\frac{1}{2^2};\frac{1}{9}=\frac{1}{3^2};\frac{1}{16}=\frac{1}{4^2};....\)

\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)( tổng 100 số hạng đầu tiên)

\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

                                                                                  \(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

                                                                                 \(=1+\left(1-\frac{1}{100}\right)=1+1-\frac{1}{100}=2-\frac{1}{100}< 2\)

\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)

100 số hạng đầu tiên của dãy là 1;1/4;1/9;...;1/10000

A=1+1/2^2+1/3^2+...+1/100^2<1+1/1.2+1/2.3+...+1/99.100=1+1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100=2-1/100<2

P/s : Lớp 6 nha bạn : 

Ta có : 

\(1=\frac{1}{1^2};\frac{1}{4}=\frac{1}{2^2};\frac{1}{9}=\frac{1}{3^2};\frac{1}{16}=\frac{1}{4^2};...\)

\(\Rightarrow\)Số hạng thứ 100 là :  \(\frac{1}{100^2}\)

Ta được dãy số : 

\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Lại có : 

\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1+1-\frac{1}{100}\)

\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 2-\frac{1}{100}\)

\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 2\left(đpcm\right)\)

~ Ủng hộ nhé 

a) Quy đồng pso và tính như bthg (4824829/6350400)

b) Vì 4814819 < 6350400 => A < 1

1/ So sánh A với \(\frac{1}{4}\)

Có \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.........+\frac{1}{2014.2015.2016}\)

\(A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-.......+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)

\(A=\frac{1}{1.2}-\frac{1}{2015.2016}=\frac{1}{2}-\frac{1}{2015.2016}\)

Vậy \(A>\frac{1}{4}\)

\(N=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}\)

\(N>\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}\)

\(N>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{10}{22}>\frac{9}{22}\)

Vậy N > 9/22 

>1/4+1/4+1/16+1/16+1/16

=2/4+3/16=11/16=66/96<80/96=5/6

vậy A<5/6

Bài 1: a)  \(M=1+5+5^2+...+5^{100}\)

\(5M=5+5^2+5^3+...+5^{101}\)

\(5M-M=\left(5+5^2+5^3+...+5^{101}\right)-\left(1+5+5^2+...+5^{100}\right)\)

\(4M=5^{101}-1\)

\(M=\frac{5^{101}-1}{4}\)

b) \(N=2+2^2+...+2^{100}\)

\(2N=2^2+2^3+...+2^{101}\)

\(2N-N=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)

\(N=2^{101}-2\)

Bài 2:

a) \(16^{32}=\left(2^4\right)^{32}=2^{128}\) 

\(32^{16}=\left(2^5\right)^{16}=2^{80}\)

Vì \(2^{128}>2^{80}\Rightarrow16^{32}>32^{16}\)

MÌNH KHÔNG HIỂU GÌ CẢ

\(a=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{4036081}\)

\(=\frac{1}{2\times2}+\frac{1}{3\times3}+\frac{1}{4\times4}+...+\frac{1}{2009\times2009}\)

\(< \frac{1}{2\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2008\times2009}\)

\(=\frac{1}{4}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+...+\frac{2009-2008}{2008\times2009}\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\)

\(=\frac{3}{4}-\frac{1}{2009}< \frac{3}{4}\)

A = 1/4 +1/9 + 1/16 + 1/25 + 1/36 
   =   ( 1/4 + 1/16 ) + ( 1/9 + 1/36) + 1/25
   =         5/16         +       5/36       + 1/25
   =                       65/144              + 1/25
   = 1769/3600
=> 1769/3600 < 5/6 (hay 1769/3600 < 3000/3600 -quyđồng-)
Vậy A< 5/6
Đúng nhé, tk cho mjk với-số to thiệt nhưng đúng mà-

\(\frac{5}{6}>A\)

A= 1/2²+1/3²+1/4²+1/5²+1/6²< 1/1x2+1/2x3+1/3x4+1/4x5+1/5x6

=> A < 1-1/6=5/6

lớp 8a3 nguyễn khuyến đúng ko