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\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(< \frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}+\frac{1}{1}=2\)
\(\Rightarrow\)\(A< 2\left(đpcm\right)\)
chúc bạn học tốt!!!
Bài 6 :
2S = 6 + 3 + 3/2 + ... + 3/2^8
2S = 6 - 3/2^9 + S
S = 6 - 3/2^9
Vậy S = 6 - 3/2^9
Bài 7 :
Ta có :
A = 1/1 + 1/2^2 + 1/3^2 + ... + 1/50^2 < 1 + 1/(1x2) + 1/(2x3) + ... + 1/(49x50) = 1 + 1 - 1/50 < 1 + 1 = 2
=) A < 2
Vậy A < 2
Bài 8 :
Do A = 1 + 2/(2015^2014 - 1 ) và B = 1 + 2/(2015^2014 - 3 ) mà 2/(2015^2014 -1) < 2/(2015^2014 - 3 )
=) A < B
Vậy A < B
Bài 9:
Do 196/197 > 196/(197+198) và 197/198 > 197/(197+198)
=) A > B
Vậy A > B
b, \(\frac{2^{10}\left(13+65\right)}{2^8.104}\)
=\(\frac{2^2.78}{104}\)=\(\frac{312}{104}\)=3
Ta có :
\(\frac{2014^{2015}+1}{2014^{2015}+1}\)\(=1\)
\(\frac{2014^{2014}+1}{2014^{2013}+1}\)\(>1\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
a = \(\frac{2013}{2014}+\frac{2014}{2015}=\frac{2014-1}{2014}+\frac{2015-1}{2015}\)
\(=1-\frac{1}{2014}+1-\frac{1}{2015}\)
\(=2-\left(\frac{1}{2014}+\frac{1}{2015}\right)>1\) (1)
b = \(\frac{2013+2014}{2014+2015}<1\) (2)
Từ (1) và (2) => a > b
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h
\(\left(\frac{5}{2014}+\frac{4}{2015}-\frac{3}{2016}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{5}{2014}+\frac{4}{2015}-\frac{3}{2016}\right).\left(\frac{1}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{5}{2014}+\frac{4}{2015}-\frac{3}{2016}\right).0=0\)
a) số số hạng
(2015-1) : 2+1=1008
tổng dãy số
1008 x (2015 +1) :2 = 1016064
a) 1+3+5+7+...+2015
Day tren co so so hang la:
(2015-1):2+1=1008(so hang)
Tong tren bang: (2015+1).1008:2=1016064
b) \(\left(2015.2014+2014.2013\right).\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)
= \(\left(2015.2014+2014.2013\right).\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)
= \(\left(2015.2014+2014.2013\right).\left(1+\frac{1}{2}.\frac{2}{3}-\frac{4}{3}\right)\)
= \(\left(2015.2014+2014.2013\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
= \(\left(2015.2014+2014.2013\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)
=\(\left(2015.2014+2014.2013\right).0\)
= \(0\)