Cho S=3/4+8/9+15/16+...+9999/10000
So sánh s với 98
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chịu mẹ kiếp toán 7 cho vào đề kiểm tra toán 6 ai mà lm dc
=1-1/4+1-1/9+1-1/16+...+1-1/10000
=(1+1+1+...+1)+(-1/4-1/9-1/16-...-1/10000)
=99+(-1/4-1/9-1/16-...-1/10000)
Vì 99+(-1/4-1/9-1/16-...-1/10000)>98
=>C>98
Vây C>98
`A=3/4+8/9+.............+9999/10000`
`=1-1/4+1-1/9+,,,,,,,,,,+1-1/10000`
`=99-(1/4+1/9+.........+1/10000)<99-0=99`
`=>A<99`
Ta có : \(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}=\frac{4-1}{4}+\frac{9-1}{9}+\frac{16-1}{16}+...+\frac{10000-1}{10000}\)
\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{100^2-1}{100^2}\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\left(99\text{ số hạng 1}\right)\)
\(=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>99-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(=99-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\right)=99-\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=99-\frac{99}{202}>99-\frac{1}{2}=98,5\)
=> A > 98,5
=> A > 98
C = ( 1 - 1/4 ) + ( 1 - 1/9 ) + ( 1 - 1/16 ) + .. .+ ( 1 - 1/10000 )
C = 1 + 1 + ... + 1 - ( 1/4 + 1/9 + 1/16 + ... + 1/10000 )
C = 1 + 1 + 1 +... + 1 - ( 1/22 + 1/32 + .. + 1/1002 )
C = 99 - ( 1/22 + 1/32 + ... + 1/1002 )
Mà 1/22 + 1/32 + ... + 1/1002 < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 = 1 - 1/2 + 1/2 - 1/3 + .. + 1/99 - 1/100 = 1 - 1/100 < 1 =>
C > 99 - 1 => C > 98
\(C=\frac{4-1}{4}+\frac{9-1}{9}+....+\frac{10000-1}{10000}.\)
\(C=1-\frac{1}{4}+1-\frac{1}{9}+.....+1-\frac{1}{10000}.\)
\(C=\left(1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+...+\frac{1}{10000}\right)\)
\(C=99-\left(\frac{1}{4}+\frac{1}{9}+...+\frac{1}{10000}\right)\)
ta có :\(\frac{1}{4}< 1,\frac{1}{9}< 1,......,\frac{1}{10000}< 1\)
\(\Rightarrow\frac{1}{4}+\frac{1}{9}+...+\frac{1}{10000}< 1\)
\(C=99-\left(\frac{1}{4}+\frac{1}{9}+...+\frac{1}{10000}\right)>98\)
vậy C>98
\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Đặt D = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
.............
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow D>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
\(\Rightarrow C=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>98\)(đpcm)
Nhanh lên cac bạn ơi
mình ra là 0