Cho biểu thức A=\(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2014}}\)Hãy so sánh A với\(\frac{3}{2}\)
Cho M=\(\frac{\sqrt{2}-\sqrt{1}}{1+1}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+...+\frac{\sqrt{2015}-\sqrt{2014}}{2014+2015}\)
Hãy so sánh M với 1/2
A=1+\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}\)
so sánh A với \(\frac{3}{2}\)
\(3A=3\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\right)\)
\(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2013}}\)
\(3A-A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\right)\)
\(2A=3-\frac{1}{3^{2014}}\)
\(A=\left(3-\frac{1}{3^{2014}}\right):2\)
\(A=\frac{3}{2}-\frac{1}{2.3^{2014}}<\frac{3}{2}\)
\(\Rightarrow A<\frac{3}{2}\)
Câu 1:
a)Cho A=\(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{5^2}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\). So sánh A với B
b) Cho B= x2013 - 2014.x2012 + 2014.x2011 - 2014.x2010 +...- 2014.x2 + 2014.x - 1. Tính giá trị biểu thức với x=2013
Bạn xem lại đề câu a) cho rõ lại
Câu b) Tại x=2013 thì B=x2013-(x+1)x2012+(x+1)x2011-(x+1)x2010+...-(x+1)x2+(x+1)x-1
= x2013-x2013-x2012+x2012+x2011-x2011-x2010+..-x3 - x2+x2+x-1
= x-1 = 2012
giá trị của biểu thức A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}}\)
Cho \(M=\frac{\sqrt{2}-\sqrt{1}}{1+2}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+\frac{\sqrt{4}-\sqrt{3}}{3+4}+...+\frac{\sqrt{2015}-\sqrt{2014}}{2014+2015}\). Hãy so sánh M với \(\frac{1}{2}\)
giá trị biểu thức A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+...+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}+\frac{1}{2015}}là?\)
Câu 5. (1,0 điểm)
Cho tổng A gồm 2014 số hạng: A = \(\frac{1}{19}+\frac{2}{19^2}+\frac{3}{19^3}+..........+\frac{2014}{19^{2014}}\)
Hãy so sánh A2013 và A2014.
Biết n!=1.2.3...n \(\left(n\inℕ^∗;n\ge2\right)\)và \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+......+\frac{2014}{2015!}\)
Hãy so sánh A với 1
Ta có \(A=\frac{1}{2!}+\frac{2}{3!}+...+\frac{2014}{2015!}\)
=> \(A=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2015-1}{2015!}\)
=> \(A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2015}{2015!}-\frac{1}{2015!}\)
=> \(A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2014!}-\frac{1}{2015!}\)
=> \(A=1-\frac{1}{2015!}< 1\)
Xét tổng T= \(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2015}{2^{2014}}\).Hãy so sánh T với 3
Ta có :
\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2015}{2^{2014}}\)
\(\frac{1}{2}T=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2015}{2^{2015}}\)
\(T-\frac{1}{2}T=\left(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2015}{2^{2014}}\right)-\left(\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2015}{2^{2015}}\right)\)
\(\frac{1}{2}T=1+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2015}{2^{2014}}-\frac{2}{2^2}-\frac{3}{2^3}-\frac{4}{2^4}-...-\frac{2015}{2^{2015}}\)
\(\frac{1}{2}T=1+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{2015}{2^{2014}}-\frac{2014}{2^{2014}}\right)-\frac{2015}{2^{2015}}\)
\(\frac{1}{2}T=1+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\right)-\frac{2015}{2^{2015}}\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^{2014}}\)
Mà \(\frac{1}{2^{2014}}>0\)
\(\Rightarrow\)\(A=\frac{1}{2}-\frac{1}{2^{2014}}< \frac{1}{2}\)
\(\Leftrightarrow\)\(1+A-\frac{2015}{2^{2015}}< 1+\frac{1}{2}-\frac{1}{2^{2014}}-\frac{2015}{2^{2015}}\)
\(\Leftrightarrow\)\(\frac{1}{2}T< \frac{3}{2}-\left(\frac{1}{2^{2014}}+\frac{2015}{2^{2015}}\right)\)
Mà \(\frac{1}{2^{2014}}+\frac{2015}{2^{2015}}>0\)
\(\Rightarrow\)\(\frac{1}{2}T< \frac{3}{2}\)
\(\Rightarrow\)\(\frac{1}{2}T.2< \frac{3}{2}.2\)
\(\Rightarrow\)\(T< 3\) ( đpcm )
Vậy \(T< 3\)
Bạn xem đúng không nhé, chúc bạn học tốt ~
Ta có : T = 2 1 2 + 2 2 3 + 2 3 4 + ... + 2 2014 2015 2 1 T = 2 2 2 + 2 3 3 + 2 4 4 + ... + 2 2015 2015 T − 2 1 T = 2 1 2 + 2 2 3 + 2 3 4 + ... + 2 2014 2015 − 2 2 2 + 2 3 3 + 2 4 4 + ... + 2 2015 2015 2 1 T = 1 + 2 2 3 + 2 3 4 + ... + 2 2014 2015 − 2 2 2 − 2 3 3 − 2 4 4 − ... − 2 2015 2015 2 1 T = 1 + 2 2 3 − 2 2 2 + 2 3 4 − 2 3 3 + ... + 2 2014 2015 − 2 2014 2014 − 2 2015 2015 2 1 T = 1 + 2 2 1 + 2 3 1 + ... + 2 2014 1 − 2 2015 2015 Đặt A = 2 2 1 + 2 3 1 + ... + 2 2014 1 2A = 2 1 + 2 2 1 + ... + 2 2013 1 2A − A = 2 1 + 2 2 1 + ... + 2 2013 1 − 2 2 1 + 2 3 1 + ... + 2 2014 1 A = 2 1 − 2 2014 1 Mà 2 2014 1 > 0 ⇒A = 2 1 − 2 2014 1 < 2 1 ⇔1 + A − 2 2015 2015 < 1 + 2 1 − 2 2014 1 − 2 2015 2015 ⇔ 2 1 T < 2 3 − 2 2014 1 + 2 2015 2015 Mà 2 2014 1 + 2 2015 2015 > 0 ⇒ 2 1 T < 2 3 ⇒ 2 1 T.2 < 2 3 .2 ⇒T < 3 ( đpcm ) Vậy T < 3 Bạn xem đúng không nhé, chúc bạn học tốt ~