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Ta có: B=1/3+2/32+3/33+...+99/399+100/3100
3B=1+1/3+2/32+3/33+...+99/399
3B-B=(1+1/3+2/32+3/33+...+99/399)-(1/3+2/32+3/33+4/34+..+99/399+100/3100)
Đặt A=1/3+1/32+1/33+..+1/399
3A=1+1/3+1/32+..+1/399
2A=1-1/399=>A=1-1/399/2
Thay vào 2B...........................
Ta sẽ ra B<3/12
-Chúc hk tốt-
A=1/3 - 2/3^2+3/3^3 - 4/3^4+ ... - 100/3^100
=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99
=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100
=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99
=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...
<=>16A=3-101/3^99-100/3^100
<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16
Suy ra A<3/16
Ta có:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
Mà \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{99}{100}\)
ta có: M = 1/3 - 2/3^2 + 3/3^3 - 4/3^4 +......+ 99/3^99 - 100/3^100
=> 3.M = 1 - 2/3 + 3/3^2 - 4/3^3 +.......+ 99/3^98 - 100/3^99
=> 3M + M = ( 1 - 2/3 + 3/3^2 - 4/3^3 +.........+ 99/3^98 - 100/3^99 ) + ( 1/3 - 2/3^2 + 3/3^3 - 4/3^4 +....+ 99/3^99 - 100/3^100 )
=> 4.M = 1- 1/3 + 1/3^2 - 1/3^3 +........+ 1/3^98 - 1/3^99 - 100/3^100
=> 12.M = 3 - 1 + 1/3 - 1/3^2 +.......+ 1/3^97 - 1/3^98 - 1/3^99
=> 12M + 4M = ( 3 - 1 + 1/3 - 1/3^2 +......+ 1/3^97 - 1/3^98 - 1/3^99 ) + ( 1 - 1/3 + 1/3^2 - 1/3^3 +.......+ 1/3^99 - 1/3^100 )
=> 16M = 3 - 101/3^99 - 100/3^100
vù 16M < 3
=> M < 3/16
vậy M < 3/16
tk cho mk nha,mk bị âm rùi