3/1^2.2^2+5/2^2.3^2+7/3^2.4^2+...+4019/2009^2.2010^2

=3/1.4+5/4.9+7/9.16+...+4019/4036081.4040100

= 1/1-1/4+1/4-1/9+1/9-1/16+...+1/4036081-1/4040100

= 1/1-1/4040100

= 1-1/4040100 < 1

Chúc bạn học tốt!

câu hỏi là tính tổng à

k bạn ơi nó là cm  cái tổng đó <1 

mấy bạn ơi câu b) là chứng minh C<\(\dfrac{1}{2}\)nha

\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{2010^2-2009^2}{2009^2.2010^2}\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2}=1-\frac{1}{2010^2}\)

\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{2010^2-2009^2}{2009^2.2010^2}\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2}=1-\frac{1}{2010^2}\)

_{^{555555+9973938782858853963.7}}

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Ta có:

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

= \(\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)

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

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

Mà \(1-\dfrac{1}{10^2}< 1\) nên:

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) < 1 (đpcm).

Ta thấy 3²-2²=5; 4²-3²=7;......;2014²-2013²=(2014-2013)(2014+2013)=4017

=> VT=1/4+1/4-1/9+1/9-1/16+1/16-......-1/2013²+1/2013²-1/2014²

=1/4+1/4-1/2014=1/2-1/2014²<1/2<1