d) Theo đề bài: ${H}$= ${2}$+$\frac{3}{4}$+$\frac{8}{9}$+$\frac{15}{16}$+.....+$\frac{2499}{2500}$
⇒ ${H}$= ${2}$ +(1-$\frac{1}{4}$)+(1-$\frac{1}{9}$)+(1-$\frac{1}{16}$)+.....+(1-$\frac{1}{2500}$)
⇒ ${H}$= ${2}$+(1-$\frac{1}{2^{2}}$)+(1-$\frac{1}{3^{2}}$)+(1-$\frac{1}{4^{2}}$)+.....+(1-$\frac{1}{50^{2}}$)
⇒ ${H}$= ${2}$+ ${50}$ - ( $\frac{1}{2^{2}}$+ $\frac{1}{3^{2}}$+$\frac{1}{4^{2}}$+...+ $\frac{1}{50^{2}}$
⇒ ${H}$= ${52}$ - ${B}$ ( ${B}$= $\frac{1}{2^{2}}$+ $\frac{1}{3^{2}}$+$\frac{1}{4^{2}}$+...+ $\frac{1}{50^{2}}$) (1)
Có : ${B}$ = $\frac{1}{2^{2}}$+ $\frac{1}{3^{2}}$+$\frac{1}{4^{2}}$+...+ $\frac{1}{50^{2}}$
Ta có:
$\frac{1}{2^{2}}$ < $\frac{1}{1.2}$
$\frac{1}{3^{2}}$ < $\frac{1}{2.3}$
...
$\frac{1}{50^{2}}$ < $\frac{1}{49.50}$
⇒${B}$ < $\frac{1}{1.2}$+$\frac{1}{2.3}$+...+$\frac{1}{49.50}$
⇒${B}$ <$\frac{1}{1}$- $\frac{1}{2}$+ $\frac{1}{2}$- $\frac{1}{3}$+....+ $\frac{1}{49}$- $\frac{1}{50}$
⇒${B}$<$\frac{1}{1}$-$\frac{1}{50}$
⇒${B}$<$\frac{49}{50}$
Từ ${(1)}$⇒ ${H}$ = ${52}$ - ${B}$> ${52}$ - $\frac{49}{50}$ > ${50}$
Vậy ${H}$> ${50}$
