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Có A = 1/2 + 1/2^2 + 1/2^3 + ......+1/2^2018
Nên 2A = 1 + 1/2 + 1/2^2 + ......+1/2^2017
Suy ra 2A - A = (1+ 1/2 + 1/2^2 +.........+1/2^2017) - (1/2 + 1/2^2 + 1/2^3 + ......+ 1/2^2^2008)
A = 1 - 1/2^2008
Nên 2^2008*A + 1 = 2^2008 * (1 - 1/2^2008) + 1
=2^2008 - 1 +1
=2^2008
Vậy, 2^2008*A+1 là 1 lũy thừa với cơ số tự nhiên
Ta có :
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)
\(2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\right)\)
\(S=1-\frac{1}{2^{2018}}\)
\(S=\frac{2^{2018}-1}{2^{2018}}\)
Vậy \(S=\frac{2^{2018}-1}{2^{2018}}\)
Chúc bạn học tốt ~
\(A=\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{2018^2}\)
\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2017\cdot2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}< \frac{3}{4}\)
Giải:
a) Đặt:
\(A=1+2^2+2^3+2^4+...+2^{2018}\)
\(\Leftrightarrow2A=2+2^3+2^4+2^5+...+2^{2019}\)
\(\Leftrightarrow2A-A=\left(2+2^{2019}\right)-\left(1+2^2\right)\)
\(\Leftrightarrow A=2+2^{2019}-1-2^2\)
\(\Leftrightarrow A=2+2^{2019}-5\)
\(\Leftrightarrow A=2^{2019}-3\)
Vậy \(A=2^{2019}-3\).
b) Đặt:
\(B=1+5+5^2+5^3+...+5^{2017}\)
\(\Leftrightarrow5B=5+5^2+5^3+5^4+...+5^{2018}\)
\(\Leftrightarrow5B-B=5^{2018}-1\)
\(\Leftrightarrow4B=5^{2018}-1\)
\(\Leftrightarrow B=\dfrac{5^{2018}-1}{4}\)
Vậy \(B=\dfrac{5^{2018}-1}{4}\).
Chúc bạn học tốt!
a)A= 1 + 22+23 + 24 +....+22018
2A = 22 + 23 + 24 +......+22018 + 22019
_
A= 1 + 22+23 + 24 +....+22018
A= 22019 - 1
\(S=2^{2019}-2^{2018}-2^{2017}-...-2^2-2-1\)
\(=2^{2019}-\left(1+2+2^2+...+2^{2017}+2^{2018}\right)\) (1)
Đặt \(Q=1+2+2^2+...+2^{2017}+2^{2018}\)
\(2Q=2+2^2+2^3+...+2^{2018}+2^{2019}\)
\(2Q-Q=2^{2019}-1\)
\(Q=2^{2019}-1\)(2)
Từ (1) và (2), ta được:
\(S=2^{2019}-\left(2^{2019}-1\right)=1\)
a)\(A=1+3+3^2+...+3^{2018}\)
\(\Rightarrow3A=3.\left(1+3+3^2+...+3^{2018}\right)\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{2019}\)
\(\Rightarrow3A-A=3+3^2+3^3+...+3^{2019}-\left(1+3+3^2+...+3^{2018}\right)\)
\(\Rightarrow2A=3^{2019}-1\)
\(\Rightarrow A=\frac{3^{2019}-1}{2}\)
b) \(B=5+5^2+...+5^{2017}\)
\(\Rightarrow5B=5^2+5^3+...+5^{2018}\)
\(\Rightarrow5B-B=5^2+5^3+...+5^{2018}-5-5^2-...-5^{2017}\)
\(\Rightarrow4B=5^{2018}-5\)
\(\Rightarrow B=\frac{5^{2018}-5}{4}\)
a,A=1+3+32+...+32017
3A=3+32+33+...+32018
3A-A=32018-1
2A=32018-1
A=(32018-1):2
\(A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2018}\)
\(2A=2+1+\frac{1}{2}+...+\left(\frac{1}{2}\right)^{2017}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\left(\frac{1}{2}\right)^{2018}\right)-\left(1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{2018}\right)\)
\(A=2-\left(\frac{1}{2}\right)^{2018}\)
ta có: \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2018}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2017}}\)
\(\Rightarrow2A-A=2-\frac{1}{2^{2018}}\)
\(A=2-\frac{1}{2^{2018}}\)