chứng minh rằng 1/3^2+1/5^2+1/7^2+…..+1/(2n+1)^2 <1/4 23/07/2021 Bởi Madelyn chứng minh rằng 1/3^2+1/5^2+1/7^2+…..+1/(2n+1)^2 <1/4
Ta có: $\frac{n}{x.(x+n)}$= $\frac{1}{x}$-$\frac{1}{x+n}$ Ta có: $\frac{1}{3²}$+$\frac{1}{5²}$+..+$\frac{1}{(2n+1)²}$ < $\frac{1}{2}$.( 1-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{5}$+…+$\frac{1}{2n-1}$-$\frac{1}{2n+1}$)= $\frac{1}{2}$.(1-$\frac{1}{2n+1}$) Với n là số tự nhiên, ta có: 2n+1> 2n > 2 ⇒ $\frac{1}{2n+1}$< $\frac{1}{2}$ ⇒ 1-$\frac{1}{2n+1}$ < $\frac{1}{2}$ ⇒ $\frac{1}{3²}$+$\frac{1}{5²}$+..+$\frac{1}{(2n+1)²}$< $\frac{1}{2}$.$\frac{1}{2}$= $\frac{1}{4}$ ( đpcm) Bình luận
Giải thích các bước giải: Ta có : $A=\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+..+\dfrac{1}{(2n+1)^2}$ $\rightarrow A<\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+..+\dfrac{1}{(2n-1).(2n+1)}$ $\rightarrow A<\dfrac{1}{2}.(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+..+\dfrac{2}{(2n-1).(2n+1)})$ $\rightarrow A<\dfrac{1}{2}.(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+..+\dfrac{1}{2n-1}-\dfrac{1}{2n+1})$ $\rightarrow A<\dfrac{1}{2}.(1-\dfrac{1}{2n+1})$ mà $2n+1\ge 2.n\ge 2\rightarrow 1-\dfrac{1}{2n+1} \ge 1-\dfrac{1}{2}=\dfrac{1}{2}$ $\rightarrow A<\dfrac{1}{4}$ Bình luận
Ta có: $\frac{n}{x.(x+n)}$= $\frac{1}{x}$-$\frac{1}{x+n}$
Ta có: $\frac{1}{3²}$+$\frac{1}{5²}$+..+$\frac{1}{(2n+1)²}$ < $\frac{1}{2}$.( 1-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{5}$+…+$\frac{1}{2n-1}$-$\frac{1}{2n+1}$)= $\frac{1}{2}$.(1-$\frac{1}{2n+1}$)
Với n là số tự nhiên, ta có: 2n+1> 2n > 2
⇒ $\frac{1}{2n+1}$< $\frac{1}{2}$
⇒ 1-$\frac{1}{2n+1}$ < $\frac{1}{2}$
⇒ $\frac{1}{3²}$+$\frac{1}{5²}$+..+$\frac{1}{(2n+1)²}$< $\frac{1}{2}$.$\frac{1}{2}$= $\frac{1}{4}$
( đpcm)
Giải thích các bước giải:
Ta có :
$A=\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+..+\dfrac{1}{(2n+1)^2}$
$\rightarrow A<\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+..+\dfrac{1}{(2n-1).(2n+1)}$
$\rightarrow A<\dfrac{1}{2}.(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+..+\dfrac{2}{(2n-1).(2n+1)})$
$\rightarrow A<\dfrac{1}{2}.(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+..+\dfrac{1}{2n-1}-\dfrac{1}{2n+1})$
$\rightarrow A<\dfrac{1}{2}.(1-\dfrac{1}{2n+1})$
mà $2n+1\ge 2.n\ge 2\rightarrow 1-\dfrac{1}{2n+1} \ge 1-\dfrac{1}{2}=\dfrac{1}{2}$
$\rightarrow A<\dfrac{1}{4}$