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. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
Chứng minh rằng \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\frac{5}{3^5}-.........+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Bài 1: Tính
a, A = 7 + 7^3 + 7^5 + ...... + 7^151
b, B = 11^4 + 11^5 + 11^6 + ....... + 11^50
c, C = ( 2/3 )^4 + ( 2/3 )^5 + ( 2/3 )^6 + ..... + ( 2/3 )^100 ( 2/3 nghĩa là 2 phần 3 )
d, D = 5^100 - 5^99 - 5^98 - 5^97 -.....- 5^2 - 5 - 1
Bài 2: Cho A = 1 + 4 + 4^2 + ..... + 4^99
B = 4^100
a, Tìm B - A
b, Chứng minh rằng A < B/3 ( B/3 nghĩa là B phần 3 )
Bài 3 : Tính
a, A = 7^2 + 14^2 + 21^2 + 28^2 + ...... + 371^2
b, B = 11^2 + 22^2 + 33^2 + ...... + 1661^2
Bài 4 : Tính
A = 99 x 1 + 98 x 2 + 97 x 3 + ....... + 3 x 97 + 2 x 98 +1 x 99
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
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\RightarrowĐPCM\)
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
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
Bài 5 chứng minh: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Tính
A=1/2+1/2^2+1/2^3+...+1/2^100
Tính
B=1/2+1/2^2+1/2^3+1/2^4+...+1/2^99 - 1/2^100
Tính
C=1/2+1/2^3+1/2^5+...+1/2^99
Tính
D=2/3+8/9+26/27+...+3^n-1/3^n.Chứng minh A>n-1/2
Tính: E=4/3+10/9+28/27+...+3^39+1/3^92.Chứng minh B<100
Tính
F=5/4+5/4^2+5/4^3+...+5/4^99.Chứng minh C<5/3
Tính
G=3/1^2*2^2+5/2^2*3^2+7/3^2*4^2+...+19/9^2*10^2.Chứng Minh D<1
a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
Chứng minh rằng \(D=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}< \frac{1}{16}\)