1 + 2 x 2 + 3 x 3 + 4 x 4 + 5 x 5 + 6 x 6 + 7 x 7 + 8 x 8 + 9 x 9 + 10 x 10

= 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100

= 385

1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7 + 8x8 + 9x9 + 10x10

= 1+4+9+16+25+36+49+64+81+100

=(81+9)+(64+16)+(49+1)+)36+4)+25+100

=90+80+50+40+25 +100

=385

1,4,9,16

ai đúng mình tích

= 4

= 9

= 16

= 25

=4

=9

=16

=25

4523526436524545765787854521215627105716579191572912815284

Ta có: \(\frac{1}{3\times3}<\frac{1}{2\times3}=\frac12-\frac13\)

\(\frac{1}{4\times4}<\frac{1}{3\times4}=\frac13-\frac14\)

...

\(\frac{1}{100\times100}<\frac{1}{99\times100}=\frac{1}{99}-\frac{1}{100}\)

Do đó: \(\frac{1}{3\times3}+\frac{1}{4\times4}+\cdots+\frac{1}{100\times100}<\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}<\frac12\)

=>\(\frac{1}{2\times2}+\frac{1}{3\times3}+\cdots+\frac{1}{100\times100}<\frac{1}{2\times2}+\frac12\)

=>\(P<\frac14+\frac12=\frac34\)

Ta có:

\(\frac{1}{2x2}<\frac{1}{1.2}\)

\(\frac{1}{3x3}<\frac{1}{2.3}\)

\(...\)

\(\frac{1}{2015x2015}<\frac{1}{2014x2015}\)

\(\Rightarrow\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{2015x2015}<\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{2014x2015}\)

\(\Rightarrow\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{2015x2015}<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(\Rightarrow\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{2015x2015}<1-\frac{1}{2015}<1\)

\(\Rightarrow\)Đpcm

\(E=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{49.49}\)

Ta có \(\frac{1}{2.2}>\frac{1}{2.3}\)

\(\frac{1}{3.3}>\frac{1}{3.4}\)

...

\(\frac{1}{49.49}>\frac{1}{49.50}\)

=> \(E=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{49.49}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{2}-\frac{1}{50}=\frac{24}{50}=\frac{12}{25}=F\)

=> E > F

Xyz olm ơi . là j vậy