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Ta có :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\)\(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=\)\(\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+\frac{4^2}{3^2.4^2}-\frac{3^2}{3^2.4^2}+...+\frac{10^2}{9^2.10^2}-\frac{9^2}{9^2.10^2}\)
\(=\)\(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=\)\(1-\frac{1}{10^2}\)
\(=\)\(\frac{100-1}{100}\)
\(=\)\(\frac{99}{100}\)
Chúc bạn học tốt ~
=3/1.4+5/4.9+7/9.16+......+19/81.100
=(1/1-1/4)+(1/4-1/9)+........+(1/81-1/100)
=1-1/100
=99/100<1(đpcm)
Chứng minh:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)\(1\)
Không khó lắm đâu, để mình. Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2+3^2}+....+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+.....+\frac{19}{81.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}<1\)
\(\frac{3}{1^22^2}+\frac{5}{2^23^2}+\frac{7}{3^24^2}+....+\frac{19}{9^210^2}< 1\)
\(A=\frac{3}{1^22^2}+\frac{5}{2^23^2}+\frac{7}{3^24^2}+....+\frac{19}{9^210^2}\)
A=\(\frac{1}{1}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
A=\(1-\frac{1}{10^2}\)
A=\(1-\frac{1}{100}\)
A=\(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^22^2}+\frac{5}{2^23^2}+\frac{7}{3^24^2}+....+\frac{19}{9^210^2}< 1\)
=\(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+.....+\frac{1}{9^2}-\frac{1}{10^2}\)
=\(1-\frac{1}{10^2}\)
Mà \(1-\frac{1}{10^2}\)\(< 1\)
=>Tổng đó bé hơn \(1\)
Mình chứng minh A<1 cho bạn nha !
A = \(\frac{3}{1.4}\)+ \(\frac{5}{4.9}\)+ .....+\(\frac{19}{81.100}\)= 1 - \(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{9}\)+ ......+ \(\frac{1}{81}\)- \(\frac{1}{100}\)= 1 - \(\frac{1}{100}\)= \(\frac{99}{100}\)< 1
Vậy A <1 (đpcm)
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+..+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}=1-\frac{1}{10^2}<1\left(đpcm\right)\)
nhớ tick nhé
a)Xét vế trái , ta có :
Gọi tổng các số hạng ở vế trái là A
=> A= \(\frac{1}{3}\)+\(\frac{1}{3^2}\)+ ... +\(\frac{1}{3^{99}}\)
=>3A = 1 + \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+ ... + \(\frac{1}{3^{98}}\)
=> 3A - A = 1 + \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+ ... + \(\frac{1}{3^{98}}\)- ( \(\frac{1}{3}\)+\(\frac{1}{3^2}\)+ ... +\(\frac{1}{3^{99}}\))
=> 2A = 1 - \(\frac{1}{3^{99}}\)
=> A = \(\frac{1}{2}\)- \(\frac{1}{3^{99}.2}\) < \(\frac{1}{2}\)
b)\(\frac{3}{1^2.2^2}\)+ \(\frac{5}{2^2.3^2}\)+ ... + \(\frac{19}{9^2.10^2}\)
= \(\frac{3}{1.4}\)+ \(\frac{5}{4.9}\)+ .... + \(\frac{19}{81.100}\)
= 1 - \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{9}\)+ ... + \(\frac{1}{81}\)- \(\frac{1}{100}\)
= 1 - \(\frac{1}{100}\) <1
a,
\(\sum\limits^{99}_{x=1}\left(\frac{1}{3^x}\right)=\frac{1}{2}\)
bài a nó có ............