Chứng tỏ rằng S=1/1!+1/2!+1/3!+...+1/2016!
Cho S=1/4+2/4^2+3/4^3+.......+2016/4^2016
Chứng tỏ rằng S<1/2
chứng tỏ rằng:
1/4<1/5+2/5^2+3/5^3+...+2016/5^2016<1/3
Đặt \(A=\frac15+\frac{2}{5^2}+\cdots+\frac{2016}{5^{2016}}\)
=>\(5A=1+\frac25+\cdots+\frac{2016}{5^{2015}}\)
=>\(5A-A=1+\frac25+\cdots+\frac{2016}{5^{2015}}-\frac15-\frac{2}{5^2}-\cdots-\frac{2016}{5^{2016}}\)
=>\(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)
Đặt \(B=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}\)
=>\(5B=1+\frac15+\cdots+\frac{1}{5^{2014}}\)
=>\(5B-B=1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac15-\frac{1}{5^2}-\cdots-\frac{1}{5^{2015}}\)
=>\(4B=1-\frac{1}{5^{2015}}=\frac{5^{2015}-1}{5^{2015}}\)
=>\(B=\frac{5^{2015}-1}{4\cdot5^{2015}}\)
TA có: \(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)
\(=1+\frac{5^{2015}-1}{4\cdot5^{2015}}-\frac{2016}{5^{2016}}=1+\frac{5^{2016}-5-8064}{4\cdot5^{2016}}=1+\frac14-\frac{8069}{4\cdot5^{2016}}\)
=>\(4A<1+\frac14=\frac54\)
=>\(A<\frac{5}{16}\)
mà \(\frac{5}{16}<\frac{5}{15}=\frac13\)
nên \(A<\frac13\) (1)
Ta có: \(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)
=>\(20A=5+1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac{2016}{5^{2015}}\)
=>\(20A-4A=5+1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac{2016}{5^{2015}}-1-\frac15-\frac{1}{5^2}-\cdots-\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)
=>\(16A=5-\frac{2017}{5^{2015}}-\frac{2016}{5^{2016}}>5\)
=>\(A>\frac{5}{16}\)
=>\(A>\frac{4}{16}=\frac14\) (2)
Từ (1),(2) suy ra 1/4<A<1/3
Cho S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2015^2}\). Chứng tỏ rằng \(\dfrac{1007}{2016}< S< \dfrac{2014}{2015}\)
Lời giải:
Ta có:
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\)
\(S> \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}\)
\(\Leftrightarrow S> \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2016-2015}{2015.2016}\)
\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(\Leftrightarrow S> \frac{1}{2}-\frac{1}{2016}=\frac{1007}{2016}\)
--------------------------
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}\)
\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{2014}{2015}\)
\(\Leftrightarrow S< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2015-2014}{2014.2015}\)
\(\Leftrightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{2014}-\frac{1}{2015}\)
\(\Leftrightarrow S< 1-\frac{1}{2015}=\frac{2014}{2015}\)
Vậy ta có đpcm.
chứng tỏ rằng B=1-1/2^2-1/3^2-1/4^2-....-1/2016^2>1/2016
Cho A=1*2*3*...*2015*2016*(1+1/2+1/3+...+1/2015+1/2016)
Chứng tỏ rằng A là số tự nhiên chia hết cho 2017
Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\) Chứng tỏ S < 1
Ta có:
S = 1/22 + 1/32 + 1/42 + ... + 1/20162
= 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/2016.2016
S < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2015.2016
S < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2015 - 1/2016
S < 1 - 1/2016
Mà 1 - 1/2016 < 1
=> S < 1
Vậy S < 1
Ủng hộ nha
Cho m/n=1+1/2+1/3+1/4+..........+1/2016
Chứng tỏ rằng m chia hết cho 2017
Cho S=1/4+2/4^2+3/4^3+...+2016/4^2016. Chứng minh rằng S<1/2
Câu 1
a) Chứng tỏ rằng 1/3 - 1/3^2 + 1/3^3 - 1/3^4 + 1/3^5 - 1/3^6 < 1/4
b) Cho A= 2015^2016 + 2016^2015 x 2015 và B= 1 + 2^2 + 3^2 + ......+2016^2. Tính AB có chia hết cho 5 không? Vì sao?