Cho S= 1/1*2+1/1*2*3+1/1*2*3*4+...+1/1*2*3*4*...*100
CT: S<1
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\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(S=\frac{1}{\frac{2}{2}}+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{n\left(n+1\right)}{2}}\)
\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{n\left(n+1\right)}\)
\(S=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(S=2.\left(1-\frac{1}{n+1}\right)< 2.1=2\)
Vậy S<2
5:
a: \(3^{2n}=\left(3^2\right)^n=9^n\)
\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)
=>\(3^{2n}>2^{3n}\)
b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)
\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)
mà \(1568239201< 8036054027\)
nên \(199^{20}< 2003^{15}\)
4: \(100< 5^{2x-1}< 5^6\)
mà \(25< 100< 125\)
nên \(125< 5^{2x-1}< 5^6\)
=>3<2x-1<6
=>4<2x<7
=>2<x<7/2
mà x nguyên
nên x=3
S = 1 + 3 + 32 + 33 + ... + 330
3S = 3 + 32 + 33 + 34 + ... + 331
3S - S = ( 3 + 32 + 33 + 34 + ... + 331 ) - ( 1 + 3 + 32 + 33 + ... + 330 )
2S = 331 - 1
S = \(\frac{3^{31}-1}{2}\)
\(S=1+3+3^2+3^3+...+3^{30}\)
\(S=1+3\left(1+3^2+...+3^{29}\right)\)
\(S=1+3\left(S-3^{30}\right)\)
\(S=1+3S-3^{31}\)
\(2S=3^{31}-1\)
\(S=\frac{3^{31}-1}{2}\)
\(N=1+4+4^2+...+4^{132}=1+4\left(1+4^2+...+4^{131}\right)\)
\(N=1+3\left(N-4^{132}\right)\)
\(N=1+3N-4^{133}=\frac{4^{133}-1}{2}\)
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)
mình ko biết dấu sao lag gì nên lam mò nhé
giả sử sao la dấu nhân
suy ra s<1/1.2+1/2.3+...+1/99.100
s<1/1-1/2+1/2-1/3+...+1/99-1/100
s<1/1-1/100
s<99/100<1
suy ra s<1
nếu sao là dấu cộng
suy ra s=+2/2.3+...+2/100.101
1/2s=1/2-1/3+1/3-1/4+...+1/100-1/101
1/2s=1/2-1/100<1/2
1/2 s <1/2 suy ra s<1
thanks ban nhiu nha