Bài làm

   \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right)\)

= \(\left(\frac{2}{2}-\frac{1}{2}\right).\left(\frac{3}{3}-\frac{1}{3}\right).\left(\frac{4}{4}-\frac{1}{4}\right).....\left(\frac{100}{100}-\frac{1}{100}\right)\)

= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)

= \(\frac{1.2.3.....99}{2.3.4.....100}\)

= \(\frac{1.1.1.....1}{1.1.1.....100}\)

= \(\frac{1}{100}\)

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cảm ơn bạn THẦN CHẾT nha!!!!!!!!

có bài nào khó thì bạn giải giúp mk nhé

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)

có phải sai để ko nhỉ

Nhan xet:

\(\frac{1}{2^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{3^2}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)

\(\frac{1}{4^2}< \frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)

....

\(\frac{1}{100^2}< \frac{1}{100.101}=\frac{1}{100}-\frac{1}{101}\)

Vay:

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{101}=\frac{99}{202}< \frac{3}{4}\)

Ta có : 

\(S=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=9.\frac{1}{10}=\frac{9}{10}\) ( vì 9 số \(\frac{1}{10}\) ) 

\(\Rightarrow\)\(S>\frac{9}{10}\)

Vậy \(S>\frac{9}{10}\)

Chúc bạn học tốt ~ 

Ta có:S= 1/2+1/3+1/4+...+1/10>1/10+1/10+..+1/10(9 số 10)

=> S> 9/10

mk nghĩ là vậy

\(S=1+2+2^2+2^3+...+2^9\)

\(2S=2+2^2+2^3+2^4+...+2^{10}\)

\(2S-S=\left(2+2^2+2^3+2^4+...+2^{10}\right)-\left(1+2+2^2+2^3+...+2^9\right)\)

\(S=2^{10}-1\)

Ta có: \(5.2^8=\left(4+1\right).2^8=4.2^8+2^8=2^2.2^8+2^8=2^{10}+2^8\)

Vậy 2¹⁰ - 1 < 2¹⁰ + 2⁸ hay S < 5.2⁸

Sửa đề: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(3S=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\ldots+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

=>3S+S=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>4S=\(1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>4A=\(-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)

=>\(A=\frac{-3^{99}-1}{4\cdot3^{99}}\)

Ta có: \(4S=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}<\frac34\)

=>\(S<\frac{3}{16}\)

mà 3/16<3/15=1/5

nên S<1/5