Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(A=1-\frac{1}{2^{2016}}< 1\)
a) M = 1 + 2 + 22 + 23 + ..... + 22019
= ( 1 + 2 + 4 ) + 23( 1 + 2 + 4 ) +.... + 22016 ( 1 + 2 + 4 )
= 7 ( 1 + 23 + 22016 ) chia hết cho 7 (đpcm)
b) M + 1 = 1 + 1 + 2 + 22 + 23 +... + 22019
= 4 + 22 + 2 3 + .....22019
= 2 x 22 + 23 + .... + 22019
= 2 x 23 + .... + 22019
= 2 x 2 2019
= 22020
Có A=1+ 1/2+1/3+... +1/2^10-1
<=> 2-1+1-1/2+1/2-1/3+...- 1/2^10-1
<=> 2-1/2^10-1
Mà 1/2^10-1 < 1 => 2-1/2^10-1 <2
=> A<10
\(A=\left(1+2\right)+2^2\left(1+2\right)+...+2^{10}\left(1+2\right)=3+2^2.3+...+2^{10}.3=3\left(1+2^2+...+2^{10}\right)⋮3\)
\(A=1+\frac{5^9}{1+5+..+5^8}\)
\(=1+\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}\)
Tương tự:
\(B=1+\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+...+\frac{1}{3}}\)
Vì \(\frac{1}{5}< \frac{1}{3}\) , \(\frac{1}{5^2}< \frac{1}{3^2}\), . . .
nên: \(\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}>\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+...+\frac{1}{3}}\)
=> A > B
Vậy đề bạn cho chứng minh A < B là sai nhé.
Ta có:\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}\)
=>\(A=\frac{\left(1+5+5^2+...+5^8\right)}{\left(1+5+5^2+...+5^8\right)}+\frac{5^9}{1+5+5^2+...+5^8}\)
=>\(A=1+\frac{5^9}{1+5+5^2+...+5^8}\)
Ta có:\(B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
=>\(B=\frac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}+\frac{3^9}{1+3+3^2+...+3^8}\)
=>\(B=1+\frac{3^9}{1+3+3^2+...+3^8}\)
vì:\(1+3+3^2+...+3^8< 1+5+5^2+...+5^8\)
Nên A<B(đpcm).
