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\(3S=1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\)
=>2S=1-1/3^100
=>S=1/2-1/2*3^100<1/2
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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)
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\(S=1+3+3^1+3^2+3^3+.....+3^{20}\)
\(3S=3.\left(1+3+3^1+3^2+3^3+.....+3^{20}\right)\)
\(3S=3.1+3.3^1+3.3^2+3.3^3+.....+3.3^{20}\)
\(3S=3+3^2+3^3+3^4+...+3^{21}\)
\(2S=3S-S\)
\(2S=\left(3+3^2+3^3+3^4+.....+3^{21}\right)-\left(1+3^1+3^2+3^3+.....+3^{20}\right)\)
\(2S=3^{21}-1\)
\(\Rightarrow S=\frac{3^{21}-1}{2}\)
\(\frac{1}{2}.3^{21}=3^{21}\div2\)
Vì \(\frac{3^{21}-1}{2}< 3^{21}\div2\)nên S < \(\frac{1}{2}.3^{21}\)
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S=1/2+1/3+....+1/63>1/30+1/30+1/30+...+1/30=1/30 x 62=31/15>2
vì có 62 số hạng nên mk nhân với 62 nha
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S = 2/2 + 2/6 + 2/12 + 2/20 + ... + 2/90
S = 2/1.2 + 2/2.3 + 2/3.4 + .. + 2/9.10
=> S = 2( 1/1.2 + 1/2.3 + ... + 1/9.10)
=> S = 2 ( 1/1 - 1/2 + 1/2 - 1/3 + .. + 1/9 - 1/10 )
=> S = 2 ( 1/1 - 1/10 )
Vì 1/1 - 1/10 < 1 => 2 ( 1/1 - 1/0 ) < 2.1 = 2
VẬy S < 2
tick đúng nha