s=1-3+3^2-3^3+...+100/3^100 hãy so sánh s với 1/5
S=1/3+2/3^2+3/3^3+4/3^4+..................+100/3^100. So sánh S với 1/5
so sánh S = 1/3 - 2/3^2 + 3/3^3 -4/3^4 + ... + 99/3^99 -100/3^100 và 1/5
Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\)
cho s=1/3-2/32+3/33-4/34+........+99/399-100/3100
so sánh s với 1/5
\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\) so sánh S với \(\dfrac{1}{2}\)
\(3S=1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\)
=>2S=1-1/3^100
=>S=1/2-1/2*3^100<1/2
Cho S=1/2^2+1/3^2+....+1/100^2 .So sánh S với 3/4
nhận xét :
\(\frac{1}{2^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3^2}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
.............
\(\frac{1}{100^2}=\frac{1}{100.101}=\frac{1}{100}-\frac{1}{101}\)
vậy
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{101}=\frac{9}{202}< \frac{3}{4}\)
Ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};.....;\frac{1}{100^2}< \frac{1}{99.100}\)
=>\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
=>\(S< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
=>\(S< \frac{1}{4}+\frac{1}{2}-\frac{1}{100}=\frac{3}{4}-\frac{1}{100}< \frac{3}{4}\)
=>S<3/4(đpcm)
ta có
1/3^2 < 1/2*3 ; 1/4^2 < 1/3*4 ; .........; 1/100^2< 1/99*100
suy ra s=1/2^2+1/3^2+....+1/100^2 < 1/2*3 + 1/3*4 +...........+ 1/99*100
S < 1/4 + 1/2 - 1/3 + 1/3 +..........+ 1/99 - 1/100
suy ra S< 1/4 +1/2 - 1/100
hay S < 3/4 -1/100
mà 3/4 -1/100< 3/4
suy ra s<3/4
So sánh S=1/2+2/2^2+3/2^+1)3+4/2^4+.....+100/2^100 với 2
Cho S= 1/3-2/32+3/33-4/34+...+99/399-100/3100. So sánh S và 1/5
Cho S=1+3^2+3^4+3^6+.......+3^98+3^100.So sánh: 8S+1 với 2^152