\(3s=3-3^2+3^3-3^4+...+3^{100}\)

\(4s=\left(3-3^2+3^3-3^4+...+3^{101}\right)+\left(1-3+3^2-3^3+...+3^{100}\right)\)

\(4s=1\)

\(s=\dfrac{1}{4}>\dfrac{1}{5}\)

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

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