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\(A=\dfrac{1}{4}+\dfrac{1}{16}+...+\dfrac{1}{576}\)
=1/2^2(1+1/2^2+...+1/12^2)
1/2^2+1/3^2+...+1/12^2<1-1/2+1/2-1/3+...+1/11-1/12=11/12
=>1+1/2^2+1/3^2+...+1/12^2<1+11/12=23/12
=>A<23/12*1/4=23/48<23/46=1/2
A = \(\dfrac{1}{4}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{1}{196}\)
A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+...+ \(\dfrac{1}{14^2}\)
A = \(\dfrac{1}{\left(1.2\right)^2}\) + \(\dfrac{1}{\left(2.2\right)^2}\) + \(\dfrac{1}{\left(2.3\right)^2}\)+...+ \(\dfrac{1}{\left(2.7\right)^2}\)
A = \(\dfrac{1}{1^2.2^2}\) + \(\dfrac{1}{2^2.2^2}\)+ \(\dfrac{1}{2^2.3^2}\)+...+ \(\dfrac{1}{2^2.7^2}\)
A = \(\dfrac{1}{2^2}\) \(\times\)( \(\dfrac{1}{1}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+...+ \(\dfrac{1}{7^2}\))
Vì \(\dfrac{1}{2}>\dfrac{1}{3}>\dfrac{1}{4}>\dfrac{1}{5}\) \(>\)\(\dfrac{1}{6}>\dfrac{1}{7}\)
⇒ \(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+\(\dfrac{1}{4.4}\)+\(\dfrac{1}{5.5}\)+\(\dfrac{1}{6.6}\)+\(\dfrac{1}{7.7}\) < \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+\(\dfrac{1}{6.7}\)
⇒ A < \(\dfrac{1}{2^2}\) \(\times\) ( 1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\))
⇒ A < \(\dfrac{1}{4}\) \(\times\) ( 2 - \(\dfrac{1}{7}\))
⇒ A < \(\dfrac{1}{2}\) - \(\dfrac{1}{28}\) < \(\dfrac{1}{2}\)
⇒ A < \(\dfrac{1}{2}\) ( đpcm)
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