Question

Chứng minh rằng:

a) P = \(\frac{12}{1.4.7}\)+\(\frac{12}{4.7.10}\)+\(\frac{12}{7.10.13}\)+...+\(\frac{12}{54.57.60}\)<\(\frac{1}{2}\)

b) S = 1+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\)<2

Answer 1

P = 2*[ 6/(1*4*7) + 6/(4*7*10) + ... + 6/(54*57*60) ]

   = 2*[ 1/(1*4) - 1/(4*7) + 1/(4*7) - 1/(7*10) + ... + 1/(54*57) -1/(57*60) ]

   = 2*[ 1/(1*4) - 1/(57*60) ]

   = 2* (427/1710)

   = 427/855 <1/2

S = 1+ 1/2^2 + 1/3^2 +... + 1/100^2

1/2^2 < 1/(1*2)

1/3^2 < 1/(2*3)

...

1/100^2 < 1/(99*100)

==> 1/2^2 +1/3^2 +.., +1/100^2 < 1/(1*2) + 1/(2*3) + ... + 1/(99*100) = 1 -1/2 +1/2 - 1/3 +1/3 -1/4 +... - 1/100

                                                                                                   =1 - 1/100 <1

==> 1/2^2 + 1/3^2 +... + 1/100^2  < 1

==> 1 + 1/2^2 + 1/3^2 +... +1/100^2 <2