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?
chứng tỏ rằng:
1/4<1/5+2/5^2+3/5^3+...+2016/5^2016<1/3
Cho A=3^1+3^2+3^3+3^4+....+3^2015+3^2016.Chứng tỏ rằng A chia hết chi 4 và 13.
\(A=3+3^2+...+3^{2016}\)
\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2015}+3^{2016}\right)\)
\(A=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+...+3^{2015}\cdot\left(1+3\right)\)
\(A=4\cdot\left(3+3^3+...+3^{2015}\right)\)
Vậy A chia hết cho 4
_____________
\(A=3+3^2+3^3+...+3^{2016}\)
\(A=\left(3+3^2+3^3\right)+...+\left(3^{2014}+3^{2015}+3^{2016}\right)\)
\(A=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+...+3^{2014}\cdot\left(1+3+9\right)\)
\(A=13\cdot\left(3+3^4+...+3^{2014}\right)\)
Vậy A chia hết cho 13
a) 125%.(-1/2)mũ 2 :(1 5/16-1,5)+2016•
b) (15 - 6 13/18):11 1/27 - 2 1/8 : 1 11/40
c) (-3,2).-15/64+(0,8 - 2 4/15): 3 2/3
d) 6/7+5/8 : 5 - 3/16 . (-2)mũ 2
e) (-5/24 + 0,75 + 7/12 ) : ( -2 1/8)
ai trả lời nhanh mik tích cho nha, cảm ơn ạ
Chứng minh rằng: a/M= 1/3-1/3^2+1/3^3-1/3^4+1/3^5-1/3^6<1/4
b/N=1/3-2/3^2+3/3^3-4/3^4+...+2015/3^2015-2016/3^2016<3/16
Cho S=1/4+2/4^2+3/4^3+.......+2016/4^2016
Chứng tỏ rằng S<1/2
a) 231 + (-54) + (-231) + (-64) + 123 + 277
b) 1 + 3 + (-5) + (-7) + 9 + 11 +...+ 97 + 99
c) 1 + (-2) + (-3) + 4 + 5 + (-6) + (-7) + 8 + 9 +...+ (-2015) + (-2016)
chứng tỏ E <3/16
E=1/3-2/3^2+3/3^3-4/3^4+...+2015/3^2015-2016/3^2016
\(E=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{2015}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 3E=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{2015}{3^{2014}}-\dfrac{2016}{3^{2015}}\\ 3E+E=\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{2015}{3^{2014}}-\dfrac{2016}{3^{2015}}\right)+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{2015}{3^{2015}}-\dfrac{2016}{3^{2016}}\right)\\ 4E=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 12E=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}-\dfrac{1}{3^{2014}}-\dfrac{6048}{3^{2016}}\\ 4E+12E=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}-\dfrac{1}{3^{2014}}-\dfrac{2016}{3^{2015}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}-\dfrac{2016}{3^{2016}}\right)\\ 16E=3-\dfrac{2017}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 16E=3-\left(\dfrac{2017}{3^{2015}}+\dfrac{672}{3^{2015}}\right)\\ 16E=3-\dfrac{2689}{3^{2015}}< 3\\ \Rightarrow E< \dfrac{3}{16}\)
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2017}}\)
chứng tỏ A<1
2,
\(S=2+2^2+2^3+...+2^{99}\)
C/t: S chia hết cho 7, 31
3,
\(A=1+5+5^2+5^3+5^4+5^5+...+5^{99}+5^{100}\)
Tính A
4,
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)<1
5,
CHỨNG tỏ rằng các p/s tối giản vs mọi số tự nhiên n
a,\(\frac{n+1}{2n+3}\)b,\(\frac{2n+3}{4n+8}\)
6,
a,TÍnh A và B
A=\(\frac{2016^{2016}+2}{2016^{1016}-1}\)B=\(\frac{2016^{2016}}{2016^{2016}-3}\)
b, tính
C=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
LÀm NHANH Hộ MK ,MAi mk Phải Nộp.
2/
S = 2 + 22 + 23 +...+ 299
= (2+22+23) +...+ (297+298+299)
= 2(1+2+22)+...+297(1+2+22)
= 2.7 +...+ 297.7
= 7(2+...+297) chia hết cho 7
S = 2+22+23+...+299
= (2+22+23+24+25)+...+(295+296+297+298+299)
= 2(1+2+22+23+24)+...+295(1+2+22+23+24)
= 2.31+...+295.31
= 31(2+...+295) chia hết cho 31
3/
A = 1+5+52+....+5100 (1)
5A = 5+52+53+...+5101 (2)
Lấy (2) - (1) ta được
4A = 5101 - 1
A = \(\frac{5^{101}-1}{4}\)
4/
Đặt A là tên của biểu thức trên
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
........
\(\frac{1}{8^2}< \frac{1}{7.8}=\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}< 1\)
Vậy...
5/
a, Gọi UCLN(n+1,2n+3) = d
Ta có : n+1 chia hết cho d => 2(n+1) chia hết cho d => 2n+2 chia hết cho d
2n+3 chia hết cho d
=> 2n+2 - (2n+3) chia hết cho d
=> -1 chia hết cho d => d = {-1;1}
Vậy...
b, Gọi UCLN(2n+3,4n+8) = d
Ta có: 2n+3 chia hết cho d => 2(2n+3) chia hết cho d => 4n+6 chia hết cho d
4n+8 chia hết cho d
=> 4n+6 - (4n+8) chia hết cho d
=> -2 chia hết cho d => d = {1;-1;2;-2}
Mà 2n+3 lẻ => d lẻ => d khác 2;-2 => d = {1;-1}
Vậy...
6/
a,Vì B > 1
\(\Rightarrow B=\frac{2016^{2016}}{2016^{2016}-3}>\frac{2016^{2016}+2}{2016^{2016}-3+2}=\frac{2016^{2016}+2}{2016^{2016}-1}=A\)
Vậy A < B
b, C = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
