CMR(1/1*2+1/2*3+1/3*4+1/4*5+...+1/99*100):(1/51+1/52+1/53+...+1/100) = 1
Sửa đề: \(\dfrac{\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
=1
Câu1: Tính tổng 100 số hạng đầu tiên của các dãy sau
a)1/3 ; 1/15 ; 1/35 ;.......
b) 1/5; 1/45 ; 1/117 ; 1/221 ;..........
Câu 2 :Rút gọn
a) P= 1+(1+2)+(1+2+3)+...+(1+2+3+...+99+100) / (1x100+2x29+...+99x2+100x1)x2013
Câu3:
Q=4+3/5+...+3/95+3/97+3/99 / 1/1.99+1/3.97+1/5.95+...+1/95.5+1/97.3+1/99.1
Câu 4 :
A)Cho S=1/1! + 1/2!+ 1/3!+...+1/2012! . Chứng Minh Rằng S<2
b) Chứng minh rằng 9/10!+10/11!+11/12!+...+99/100! < 1/9!
Câu 5
a) Cho P=1-1/2+1/3-1/4+...+1/2011-1/2012
Q=1/1007+1/1008+..+1/2011+1/2012. Tính P: Q
b) Cho M= 1/2 - 3/4 + 5/6 - 7/8 +...+ 197/198 - 199/200 và N = 1/51+1/52+1/53+...+1/100
Tính N : M
Thanks bạn Đinh Tuấn Việt nhiều nah!!!!
1/ cmr:1/2*5+1/5*8+...+1/(3n-1)(3n-2)=n/6n+4
2/ tính:
A=1/2-1/2^2+1/2^3-1/2^4+1/2^5-...+1/2^99-1/2^100
B=(1/2+1)(1/3+1)(1/4+1)...(1/99+1)
Chứng minh rằng :
a,1- 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...... + 1/ 99 - 1/ 100 = 1 / 51 + 1/ 52 + 1/ 53 + ... + 1/ 100
b, A= 1/3 - 2/ 32 + 3/ 33 - 4/ 34 + .... + 99/ 399 - 100/ 3100 < 3/ 16
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\RightarrowĐPCM\)
1) Tính: A= 2/4.7-3/5.9+2/7.10-3/9.13+..+2/301.304-3/401.405
2) Chứng minh rằng với mọi n thuộc số tự nhiên, n lớn hơn hoặc bằng 2: 3/9.14+3/14.19+...+3/(5n-1).(5n+4)<1/15
3) a) Cho A=9/5^2+9/11^2+9/17^2+...+9/305^2. Chứng minh A<3/4
b) Cho C=4/3+7/3^2+10/3^3+...+3n+1/3^n với số tự nhiên khác 0. Chứng minh rằng C<11/4
4) Tính: a) =1/2+1/2^2+1/2^3+...+1/2^100
b) B=1/3-1/3^2+1/3^3-1/3^4+...+1/3^99-1/3^100
5) So sánh: (1-1/2).(1-1/3).(1-1/4). ... .(1-1/20) với 1/21
tính tổng;
a,s=1+2+3+4+....+100
b,s=1+2+3+....+n
c,a=1+3+5+....+99
d,b=2+4+6+....+100
e,c=1+3+5+...+[2n+1][n thuộc n*]
f,d=2+4+6+...+2n
giúp tôi với
1 + 2 + 3 + ... + 100
= (100 + 1).100 : 2
= 101.50
= 5050
a) \(S=1+2+3+4+...+100\)
\(S=\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}\)
\(S=5050\)
b) \(S=1+2+3+...+n\)
\(S=\frac{\left(n+1\right)\left[\left(n-1\right):1+1\right]}{2}\)
c) \(A=1+3+5+...+99\)
\(A=\frac{\left(99+1\right)\left[\left(99-1\right):2+1\right]}{2}\)
\(A=2500\)
Chứng minh rằng:
a) A=1/2+2/2^2+3/2^3+4/4^4+...+100/3^100<2
b) B=1/3+2/3^2+3/3^3+...+100/3^100<3/4
c) C=1/2^3+1/3^3+1/4^3+...+1/n^3<1/4 (n thuộc N; n> hoặc = 2)
d) D=1/3^3+1/4^3+1/5^3+...+1/n^3<1/12 (n thuộc N; n> hoặc =3)
e) E=2/1*4/3*6/5*...*200/199<20
f) F=3/4+5/56+7/144+...+2n+1/n^2+(n+1)^2 ( n nguyên dương)
g) G=1/2*(1/6+1/24+1/60+...+1/9240)>57/62
h) H=1/31+1/32+1/33+...+1/2048>3
i) I=(1-1/3)*(1-1/6)*(1-1/10)*...*(1-1/253)<2/5
j) J=1/2!+2/3!+3/4!+...+n-1/n!<2
k) K=1/2!+5/3!+11/4!+...+n^2+n-1/(n+1)!<2 (n nguyên dương)
l) 1/6<L=1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
a/(Sửa đề bài) A= 1/2 + 2/22 + 3/23 + 4/24 +..+ 100/2100 => 1/2A = 1/22 + 2/23 + 3/24 +..+ 100/2101 => A - 1/2A = 1/2 + 2/22 +..+ 100/2100 - 1/22 - 2/23 -..- 100/2101 => 1/2A = 1/2 + 1/22 + 1/23 +..+ 1/2100 - 100/2101 Gọi riêng cụm (1/2 + 1/22 +..+ 1/2100) là B => 2B = 1 + 1/2 + 1/22 +..+ 1/299 => 2B-B = B = 1+ 1/2 +1/22 +..+ 1/299 - 1/2 - 1/22 -..- 1/2100 = 1 - 1/2100 => 1/2A = 1 - 1/2100 - 100/2101 Có 1/2A < 1 => A < 2 =>ĐPCM b/ => 1/3C = 1/32 + 2/33 + 3/34 +..+ 100/3101 => C - 1/3C = 2/3C = 1/3 + 2/32 +..+ 100/3100 - 1/32 - 2/33 -..- 100/3101 = 1/3 + 1/32 + 1/33 +..+ 1/3100 - 100/3101 Gọi riêng cụm (1/3 + 1/32 +..+ 1/3100) là D => 3D = 1 + 1/3 +..+ 1/399 => 3D - D = 2D = 1 + 1/3 +..+1/399 - 1/3 -1/32 -..- 1/3100 = 1 - 1/3100 => 2/3C *2 = 4/3C = 1 - 1/3100 - 200/3101 Có 4/3C < 1 => C<3/4 => ĐPCM Tạm thời thế đã, giải tiếp đc con nào mình sẽ gửi sau :)
Chứng minh rằng :
a,1- 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...... + 1/ 99 - 1/ 100 = 1 / 51 + 1/ 52 + 1/ 53 + ... + 1/ 100
b, A= 1/3 - 2/ 32 + 3/ 33 - 4/ 34 + .... + 99/ 399 - 100/ 3100 < 3/ 16
Giup tui nha ... Lam on ma
(1/51+1/52+1/53+...+1/100)÷(1/1×2+1/3×4+1/4×5+...+1/99×100)
