Chứng minh : $\frac{1}{2^{2}}$ + $\frac{1}{4^{2}}$ + $\frac{1}{6^{2}}$ + … +$\frac{1}{100^{2}}$ < $\frac{1}{2}$ 15/10/2021 Bởi Valerie Chứng minh : $\frac{1}{2^{2}}$ + $\frac{1}{4^{2}}$ + $\frac{1}{6^{2}}$ + … +$\frac{1}{100^{2}}$ < $\frac{1}{2}$
Giải thích các bước giải: Ta có: $A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+…+\dfrac{1}{100^2}$ $\to 2^2A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+…+\dfrac{1}{50^2}$ $\to 2^2A<\dfrac{1}{1}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{49.50}$ $\to 2^2A<1+\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+…+\dfrac{50-49}{49.50}$ $\to 2^2A<1+\dfrac11-\dfrac12+\dfrac12-\dfrac13+…+\dfrac1{49}-\dfrac1{50}$ $\to 2^2A<2-\dfrac1{50}$ $\to 2^2A<2$ $\to A<\dfrac2{2^2}$ $\to A<\dfrac12$ Bình luận
Giải thích các bước giải:
Ta có:
$A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+…+\dfrac{1}{100^2}$
$\to 2^2A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+…+\dfrac{1}{50^2}$
$\to 2^2A<\dfrac{1}{1}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{49.50}$
$\to 2^2A<1+\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+…+\dfrac{50-49}{49.50}$
$\to 2^2A<1+\dfrac11-\dfrac12+\dfrac12-\dfrac13+…+\dfrac1{49}-\dfrac1{50}$
$\to 2^2A<2-\dfrac1{50}$
$\to 2^2A<2$
$\to A<\dfrac2{2^2}$
$\to A<\dfrac12$