Đáp án: C < $\frac{3}{2}$
Giải thích các bước giải:
Ta có:
C = $\frac{1}{20}$ + $\frac{1}{21}$ + $\frac{1}{22}$ + ... + $\frac{1}{59}$
= ($\frac{1}{20}$ + $\frac{1}{21}$ + $\frac{1}{22}$ + ... + $\frac{1}{29}$) + ($\frac{1}{30}$ + $\frac{1}{31}$ + $\frac{1}{32}$ + ... + $\frac{1}{59}$)
Vì $\frac{1}{21}$ < $\frac{1}{20}$;
$\frac{1}{22}$ < $\frac{1}{20}$;
...
$\frac{1}{29}$ < $\frac{1}{20}$;
và $\frac{1}{31}$ < $\frac{1}{30}$;
$\frac{1}{32}$ < $\frac{1}{30}$;
...
$\frac{1}{59}$ < $\frac{1}{30}$;
nên C = ($\frac{1}{20}$ + $\frac{1}{21}$ + $\frac{1}{22}$ + ... + $\frac{1}{29}$) + ($\frac{1}{30}$ + $\frac{1}{31}$ + $\frac{1}{32}$ + ... + $\frac{1}{59}$) < $\frac{1}{20}$.10 + $\frac{1}{30}$.30 = $\frac{3}{2}$
