1/2^2+1/4^2+1/6^2+…+1/20^2 . Chứng minh A<1/2 13/08/2021 Bởi Arianna 1/2^2+1/4^2+1/6^2+…+1/20^2 . Chứng minh A<1/2
GIẢI A=$\frac{1}{2²}$.(1+$\frac{1}{2²}$ +$\frac{1}{3²}$+…+ $\frac{1}{10²}$) Ta có: $\frac{1}{2²}$ < $\frac{1}{1.2}$ $\frac{1}{3²}$ < $\frac{1}{2.3}$ … $\frac{1}{10²}$ < $\frac{1}{9.10}$ ⇒$\frac{1}{2²}$ +$\frac{1}{3²}$+…- $\frac{1}{10²}$ < $\frac{1}{1.2}$ + $\frac{1}{2.3}$ +… + $\frac{1}{9.10}$ = 1 – $\frac{1}{2}$ + $\frac{1}{2}$ – $\frac{1}{3}$ + … – $\frac{1}{10}$ = 1 – $\frac{1}{10}$ ⇒ $\frac{1}{2²}$ +$\frac{1}{3²}$+…+ $\frac{1}{20²}$ < 1 ⇒ 1 + $\frac{1}{2²}$ +$\frac{1}{3²}$+…+ $\frac{1}{10²}$ + 1 + 1 = 2 ⇒ A=$\frac{1}{2²}$.(1+$\frac{1}{2²}$ +$\frac{1}{3²}$+…+ $\frac{1}{10²}$) < $\frac{1}{2²}$ . 2 = $\frac{1}{2}$ (đpcm) cs gì sai sót bỏ qua giúp mk nha Bình luận
Đáp án: $A<\dfrac{1}{2}$ Giải thích các bước giải: $A=\dfrac{1}{2^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{6^{2}}+…+\dfrac{1}{20^{2}}$$=\dfrac{1}{2^{2}}\left ( 1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{10^{2}} \right )$Đặt $B=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{10^{2}}$Ta có: $\dfrac{1}{2^{2}}<\dfrac{1}{1.2}$ $\dfrac{1}{3^{2}}<\dfrac{1}{2.3}$ $…$ $\dfrac{1}{10^{2}}<\dfrac{1}{9.10}$$\Rightarrow \dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{10^{2}}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{9.10}$$\Leftrightarrow B<1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{9.10}$$\Leftrightarrow B<1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+…+\dfrac{1}{9}-\dfrac{1}{10}$$\Leftrightarrow B<2-\dfrac{1}{10}<2$$\Leftrightarrow A<\dfrac{1}{2^{2}}.2=\dfrac{1}{2}$ Bình luận
GIẢI
A=$\frac{1}{2²}$.(1+$\frac{1}{2²}$ +$\frac{1}{3²}$+…+ $\frac{1}{10²}$)
Ta có:
$\frac{1}{2²}$ < $\frac{1}{1.2}$
$\frac{1}{3²}$ < $\frac{1}{2.3}$
…
$\frac{1}{10²}$ < $\frac{1}{9.10}$
⇒$\frac{1}{2²}$ +$\frac{1}{3²}$+…- $\frac{1}{10²}$ < $\frac{1}{1.2}$ + $\frac{1}{2.3}$ +… + $\frac{1}{9.10}$ = 1 – $\frac{1}{2}$ + $\frac{1}{2}$ – $\frac{1}{3}$ + … – $\frac{1}{10}$ = 1 – $\frac{1}{10}$
⇒ $\frac{1}{2²}$ +$\frac{1}{3²}$+…+ $\frac{1}{20²}$ < 1
⇒ 1 + $\frac{1}{2²}$ +$\frac{1}{3²}$+…+ $\frac{1}{10²}$ + 1 + 1 = 2
⇒ A=$\frac{1}{2²}$.(1+$\frac{1}{2²}$ +$\frac{1}{3²}$+…+ $\frac{1}{10²}$) < $\frac{1}{2²}$ . 2 = $\frac{1}{2}$ (đpcm)
cs gì sai sót bỏ qua giúp mk nha
Đáp án: $A<\dfrac{1}{2}$
Giải thích các bước giải:
$A=\dfrac{1}{2^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{6^{2}}+…+\dfrac{1}{20^{2}}$
$=\dfrac{1}{2^{2}}\left ( 1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{10^{2}} \right )$
Đặt $B=1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{10^{2}}$
Ta có: $\dfrac{1}{2^{2}}<\dfrac{1}{1.2}$
$\dfrac{1}{3^{2}}<\dfrac{1}{2.3}$
$…$
$\dfrac{1}{10^{2}}<\dfrac{1}{9.10}$
$\Rightarrow \dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{10^{2}}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{9.10}$
$\Leftrightarrow B<1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{9.10}$
$\Leftrightarrow B<1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+…+\dfrac{1}{9}-\dfrac{1}{10}$
$\Leftrightarrow B<2-\dfrac{1}{10}<2$
$\Leftrightarrow A<\dfrac{1}{2^{2}}.2=\dfrac{1}{2}$