SO SÁNH A với 1/2
A= 1/3+1/3^2+1/3^3+...+1/3^99
So sánh:
A= 1/3 + 1/3^2 + 1/3^3 +...+ 1/3^99 với 1/2
Nhầm
\(A=\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^{99}}\)
\(\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{100}}\)
\(A-\frac{1}{3}A=\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+\left(\frac{1}{3^3}-\frac{1}{3^3}\right)+......+\left(\frac{1}{3}-\frac{1}{3^{100}}\right)\)
\(\frac{2}{3}A=\frac{1}{3}-\frac{1}{3^{100}}
So sánh A= 1/3+1/3^2+1/3^3+..+1/3^99 với 1/2
Help me!!!
A= 1/3+1/3^2+1/3^3+...+1/3^99
=> 3A=1+1/3+1/3^2+...+1/3^98
Vậy 2A= 3A-A= 1-1/3^99
=> A= 1/2 -1/ 2.3^99
=> A < 1/2
Mik giải ngắn gọn thôi nha!
a,Cho B = 1/2+1/2^2+1/2^3+...+1/2^99. So sánh B với 1
b, Cho C = 1/3+(1/3)^2+(1/3)^2+(1/3)^3+...+(1/3)^99. CMR C < 1/2
a,Cho B = 1/2+1/2^2+1/2^3+...+1/2^99. So sánh B với 1
b, Cho C = 1/3+(1/3)^2+(1/3)^2+(1/3)^3+...+(1/3)^99. CMR C < 1/2
ta có: 2B=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)
B=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+..+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
=>2B-B=\(1-\frac{1}{2^{99}}\)
mà 1/2^99>0 nên B<1 (đpcm)
A = 1/3 + 1/32 + 1/33 + ... + 1/399
So sánh A với 1/2
3A= 1+ 1/3 + 1/3^2 + ... + 1/3^98
3A-A=1 + 1/3 + 1/3^2 + ... + 1/3^98 - 1/3 - 1/3^2 - 1/3^3 - .... - 1/3^99
2A= 1 - 1/3^99 < 1
=> A < 1/2
So sánh :
A = 1/3 + 1/32 + 1/33 + ... + 1/399 với 1/2
3A = 1+1/3+1/3^2+...+1/3^99
3A-A=(1+1/3+...+1/3^99)-(1/3+1/3^2+...+1/3^99)
2A= 1-1/3^99
A = (1-1/3^99)/2 < 1/2
=> A < 1/2
so sánh
a) A=1/31+1/32+1/33+.....+1/399+1/3100 với 1/2
So sánh a và b biết :
\(\dfrac{-1}{2}\)\(-\)\(\dfrac{3-2a}{3}\)>\(\dfrac{-1}{2}\)\(-\dfrac{3-2a}{3}\)
A=1+1/2+2/3+3/4+......+99/100 so sánh với B=100-(1/2+1/3+1/4+......+1/100)
giúp mik với
ta có
\(B=1+\left(1-\frac{1}{2}\right)+..+\left(1-\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=A\)
Vậy A=B