Chứng minh (1/2!+2/3!+3/4!+4/5!+5/6!+...+99/100!) <1
Chứng minh : 7/12 < 1/ 1×2 + 1/ 3×4 + 1/ 5×6 + ........ + 1/ 99 ×100 < 5/6
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)
Ta có:\(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75};\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)
Tự giải tiếp hay nhờ thầy cô giảng tiếp đi nha bn, mỏi tay nên ko thể làm đc nữa !!
Cho M=1/2. 3/4 .5/6 . ..........99/100; N=2/3 . 4/5 . 6.7. .......100/101
a)chứng minh M<N
b)tìm tích M.N
c)chứng minh M<1/10
M=1/2*3/4*5/6*....*99/100
N=2/3*4/5*6/7*...*100/101
a, chứng minh rằng: M<N
b, tính M*N
c, chứng minh rằng: M<1/10
bài 1
A=1*2*3+2*3*4+3*4*5+...+99*100*101
B=1*3*5+3*5*7+...+95*97*99
C=2*4+4*6+..+98*100
D=1*2+3*4+5*6+...+99*100
E=1^2+2^2+3^2+...+100^2
G=1*3+2*4+3*5+4*6+...+99*101+100*102
H=1*2^2+2*3^2+3*4^2+...+99*100^2
I=1*2*3+3*4*5+5*6*7+7*8*9+...+98*99*100
K=1^2+3^2+5^2+...+99^2
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
Cho A = (1:2).(3:4).(5:60) ..... (99:100)
Cho B = (2:3).(4:5).(6:7) ..... (100:101)
Chứng minh A<B
Tính tích A , B
Chứng minh A < 1:10
a, ta xét:
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
.....
\(\frac{99}{100}< \frac{100}{101}\)
=>\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{100}{101}\)
hay:A<B(đpcm)
b,\(A.B=\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}.\frac{2}{3}.\frac{4}{5}.....\frac{100}{101}\)
\(=\frac{1.2.3....100}{2.3.4....101}=\frac{1}{101}\)
c,vì A<B (theo phần a)
=>A.A<B.A
Mà B.A=\(\frac{1}{101}\)
=>A2<101
Mà A2=\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)
=>\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)<\(\frac{1}{101}\)<\(\frac{1}{100}=\frac{1}{10^2}\)
=>\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)<\(\frac{1}{10^2}\)
=>\(\frac{1}{2}.\frac{3}{4}....\frac{99}{100}< \frac{1}{10}\)
Hay A<\(\frac{1}{10}\)
A=1/2^2+1/100^2 Chứng minh rằng A<1
B=1/1^2+1/1^2+1/3^2+...+1/100^2 Chứng minh rằng B<1 3/4 (hỗn số nhé)
C=1/1^2+1/4^2+1/6^2+...+1/100^2 Chứng minh rằng C<1/2
D=1/4^2+1/5^2+1/6^2+...+1/99^2+1/100^2 Chứng minh rằng 1/5<D<1/3
Giup mình nha mình đang cần gấp
a>
\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000
ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )
1/100^2<1/2
=>A<1
Bài 1 : Chứng minh rằng :
a) 4/3 + 4/3^2 + 4/3^3 + ..... + 4/3^99 < 2
b) B = 1/5 + 2/5^2 + 3/5^3 + ..... + 100/5^100 < 5/16
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
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\RightarrowĐPCM\)