Chứng minh rằng: A=1/4+1/16+1/36+1/64+...+1/576<1/2
Chứng tỏ rằng:A=1/4+1/16+1/36+1/64+...+1/576<1/2
THAM KHẢO NHA
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chứng minh rằng a 1/4 +1/16+1/36+1/64+1/100+1/144 +1/196+......+1/10000 <1/2
Cho A=1/4+1/16+1/36+1/64+1/100+1/144+1/196. Chứng minh rằng A < 1/2
dpcm là điều phải chứng minh nha
Ta có : \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
\(\frac{1}{16}< \frac{1}{4}-\frac{4}{8}\)
\(\frac{1}{36}< \frac{1}{8}-\frac{1}{12}\)
\(\frac{1}{64}< \frac{1}{12}-\frac{1}{16}\)
\(\frac{1}{100}< \frac{1}{16}-\frac{1}{20}\)
\(\frac{1}{144}< \frac{1}{20}-\frac{1}{24}\)
\(\frac{1}{196}< \frac{1}{24}-\frac{1}{28}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{28}\)
\(=\frac{1}{2}-\frac{1}{28}< \frac{1}{2}\)
Vậy A<1/12
Chứng minh rằng A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1324 < 1/2
hình như phân số cuối phải là 1/324
nếu là 1/324 thì tớ giải nè:
A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324
= 1/4.(1+1/2^2+1/3^2+1/4^2+1/5^2+1/6^2+1/7^2+1/8^2+1/9^2) <1/4.(1+1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)
= 1/4.(1+1-1/9)
= 1/4.17/9 = 17/36<18/36 = 1/2
=> A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324<1/2
cho dung na
nha bai tren sai day yhemh moi dung ne
Chứng minh rằng: A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324<1/2
A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324
= 1/4.(1+1/2^2+1/3^2+1/4^2+1/5^2+1/6^2+1/7^2+1/8^2+1/9^2) <1/4.(1+1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)
= 1/4.(1+1-1/9)
= 1/4.17/9 = 17/36<18/36 = 1/2
=> A = 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324<1/2
Chứng minh rằng : \(\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+........+\dfrac{1}{2020^2}< \dfrac{1}{4}\)
\(A=\dfrac{1}{4}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{1010^2}\right)\)
1/2^2+1/3^2+...+1/2010^2<1/1*2+1/2*3+...+1/2009*2010=1-1/2010<1
=>A<1/4
chứng minh rằng
1/4+1/16+1/36+1/64+1/100+1/144+1/196<1/2
khó hiểu lên thông cảm
P = 1/4 + 1/16 + 1/36 + .. + 1/196 = 1/2² + 1/4² + 1/6² +...+ 1/12² + 1/14²
xét tổng quát với số nguyên dương k ta có:
(2k-1)(2k+1) = 4k² - 1 < 4k² = (2k)² => 1/(2k)² < 1/(2k-1)(2k+1)
=> 2/(2k)² < 2 /(2k-1)(2k+1) = 1/(2k-1) - 1/(2k+1) (*)
ad (*) cho k từ 1 đến 7
2/2² < 1/1 - 1/3
2/4² < 1/3 - 1/5
...
2/12² < 1/11 - 1/13
2/14² < 1/13 - 1/15
+ + cộng lại + +
2/2² + 2/4² +...+ 2/14² < 1/1 - 1/15 < 1
=> 2(1/2² + 1/4² +..+ 1/14²) < 1 => P < 1/2 (đpcm)
Chứng minh rằng: 1/4 + 1/16 + 1/36 + 1/64 + 1/144 + 1/19 < 1/2
bài này mình làm rùi nhưng bây giờ không nhớ ,
Chứng minh rằng 1/4 + 1/16 + 1/36 + 1/64 + 1/100 + 1/144 +1/196 + ... + 1/10. 000<1/2
ta có: \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}+...+\frac{1}{10000}\)
\(=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}+...+\frac{1}{100^2}\)
Lại có: \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+...+\frac{1}{100^2}< \frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+...+\frac{1}{98.100}\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+...+\frac{1}{100^2}< \frac{1}{2^2}+\frac{49}{200}=\frac{99}{200}< \frac{100}{200}< \frac{1}{2}\)
=> đ p c m