Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
=> \(4.S=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\)
=> 4.S - S = \(\left(1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\right)\)
=> 3.S = \(=1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+\left(\frac{4}{4^3}-\frac{3}{4^3}\right)+...+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)
=> 3.S = \(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
Tính A= \(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)
=> \(4.A=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}\)
=> 4.A - A = \(4-\frac{1}{4^{2013}}\)=> A= \(\frac{4}{3}-\frac{1}{3.4^{2013}}\)
=> 3.S = \(\frac{4}{3}-\frac{1}{3.4^{2013}}-\frac{2014}{4^{2014}}\) => S = \(\frac{4}{9}-\frac{1}{9.4^{2013}}-\frac{2014}{4^{2014}}<\frac{4}{9}<1\)=> S < 1 => đpcm
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{10^2}\)
\(D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{9.10}\)
\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(D< 1-\dfrac{1}{10}\Leftrightarrow D< 1\left(đpcm\right)\)
a/M=2/3.5+2/5.7+2/7.9+.....+2/97.99
M=1/3-1/5+1/5-1/7+..+1/97-1/99
M=1/3-1/99
M=32/99
b)ta có 1/2.3+1/3.4+1/4.5+..+1/2015.2016+1/2016.2017<A
=>1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016+1/2016-1/2017<a
1/2-1/2017<A
2/15/4034<A (1)
Ta có
1/1.2+1/2.3+1/3.4+1/4.5+..+1/2015.2016>A
=>1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016>A
1-1/2016
2015/2016>A (2)
Từ (1) và (2)=>A không phải là số tự nhiên(đpcm)
Bài 1:
a) Đặt A = 1 + 7 + 72 + 73 + ... + 72016
7A = 7 + 72 + 73 + 74 + ... + 72017
7A - A = (7 + 72 + 73 + 74 + ... + 72017) - (1 + 7 + 72 + 73 + ... + 72016)
6A = 72017 - 1
\(A=\frac{7^{2017}-1}{6}\)
b) Đặt B = 1 + 4 + 42 + 43 + ... + 42017
4B = 4 + 42 + 43 + 44 + ... + 42018
4B - B = (4 + 42 + 43 + 44 + ... + 42018) - (1 + 4 + 42 + 43 + ... + 42017)
3B = 42018 - 1
\(B=\frac{4^{2018}-1}{3}\)
Bài 2:
a) Ta có: \(14\equiv1\left(mod13\right)\)
\(\Rightarrow14^{14}\equiv1\left(mod13\right)\)
\(\Rightarrow14^{14}-1⋮13\left(đpcm\right)\)
b) Ta có: \(2015\equiv1\left(mod2014\right)\)
\(\Rightarrow2015^{2015}\equiv1\left(mod2014\right)\)
\(\Rightarrow2015^{2015}-1⋮2014\left(đpcm\right)\)
Sorry mình thiếu 1+7+72+73+...+72016 câu dưới cũng thiếu 4 nha