Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = 1 + 2 + 2^2 + 2^3 + ... + 2^100
2A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^101
2A - A = A = ( 2 + 2^2 + 2^3 + 2^4 + ... + 2^101 ) - ( 1 + 2 + 2^2 + 2^3 + ... + 2^100 )
A = 2^101 - 1
Vì 2^101 - 1 < 2^101 nên A < B hay B > A
Ta có:
\(A=2^0+2^1+2^2+2^3+...+2^{100}\)
\(A=1+2+2^2+2^3+...+2^{100}\)
\(2A=2+2^2+2^3+2^4+...+2^{101}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{101}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)
\(A=2^{101}-1\left(1\right)\)
\(B=2^{101}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)suy ra:\(A< B\)
Vậy \(A< B\)
CHÚC BN HOK TỐT NHA
\(3A=\frac{1}{1}+\frac{2}{3}+\frac{3}{3^2}+....+\frac{101}{3^{100}}\)
\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}-\frac{101}{3^{101}}< 1+\frac{1}{3}+...+\frac{1}{3^{100}}< \frac{3}{2}\Rightarrow A< \frac{3}{4}\)
chủ yếu là cách làm thôi, có gì bạn tự tính
Ta có:
\(3D=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3D-D=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2D=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
Đặt \(E=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3E=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3E-E=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2E=3-\frac{1}{3^{99}}< 3\)
\(E< \frac{3}{2}\)
\(2D< \frac{3}{2}-\frac{1}{3^{100}}< \frac{3}{2}\)
\(D< \frac{3}{4}\)
Vậy...
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)
\(2A=1+\left(\frac{1-\frac{1}{3^{100}}}{2}\right)-\frac{101}{3^{101}}< 1+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow A< \frac{3}{2}:2=\frac{3}{4}\)( đpcm )
Giải :
a, Ta có :
2150 = (23)50 = 850 (1)
Lại có :
3100 = (32)50 = 950 (2)
Từ (1) và (2) => 2150 < 3100 (vì 850 < 950 )
b, Ta có :
224 = (23)8 = 88 (1)
Lại có :
316 = (32)8 = 98 (2)
Từ (1) và (2) => 224 < 316 (vì 88 < 98 )
2150=(23)50=850 < 950=(32)50=3100
224=(23)8=88 < 98 =(32)8=316
Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(3A=3\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A-A=2A\)
\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{1}{3^1}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(=1-\frac{1}{3^{100}}\)
\(2A=1-\frac{1}{3^{100}}\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
De minh lam cho:
Ta co: 2^150=(2^3)^50=8^50
3^100=(3^2)^50=9^50
Vi 8^50 < 9^50 =>2^150 < 3^100
Chuc ban hoc tot!
2^150 và 3^100
ta co: 2^150 =2^3*50=(2^3)^50=8^50
3^100=3^2.100=(3^2)^50=9^50
vì 8^50<9^50 suy ra 2^150<3^100
2^45 và 3^30
Ta có : 2^45=2^3*15=(2^3)^15=8^15
3^30=3^2*15(3^2)^15=9^15
Vì 8^15<9^15 suy ra 2^45<3^30
\(S=3+3^2+3^3+....+3^{100}\)
\(3S=3^2+3^3+3^4+...+3^{101}\)
\(3S-S=3^2+3^3+3^4+...+3^{101}-3-3^2-3^3-....-3^{100}\)
\(2S=3^{101}-3\)
\(S=\frac{3^{101}-3}{2}\)
Mà \(P=3^{101}\)
=> S < P
Mình sửa lại đề là P = 3101 nhé, chứ ko để 2101 thì ko làm được