\(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{200^2}\)
\(A=\dfrac{1}{\left(3+4+5+...+200\right)^2}\)
\(A=\dfrac{1}{\left(200-3+1\right)^2}\)
\(A=\dfrac{1}{198^2}\)
\(A=\dfrac{1\cdot198}{198}\)
\(A=\dfrac{1\cdot2\cdot9\cdot11}{2\cdot3\cdot11}\)
\(A=\dfrac{1\cdot3}{9}\)
\(A=\dfrac{1}{3}\)
Vậy A <1
Có: \(\left\{{}\begin{matrix}\dfrac{1}{3^2}< \dfrac{1}{2.3}\\\dfrac{1}{4^2}< \dfrac{1}{3.4}\\...\\\dfrac{1}{200^2}< \dfrac{1}{199.200}\end{matrix}\right.\)
Cộng vế theo vế, ta được:
\(A< \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{199.200}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}=\dfrac{1}{2}-\dfrac{1}{200}< \dfrac{1}{2}+\dfrac{1}{2}=1\left(đpcm\right)\)
