Tính
K = 1+5+5^2+...+5^2018
K
Khách
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Những câu hỏi liên quan
TV
2
31 tháng 7 2018
\(3A=1+\frac{1}{3}+...+\frac{1}{3^{2017}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{2017}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\right)\)
\(2A=1-\frac{1}{3^{2018}}\)
\(A=\frac{1-\frac{1}{3^{2018}}}{2}\)
31 tháng 7 2018
đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\right)\)
\(2A=1-\frac{1}{3^{2018}}\)
\(A=\frac{1-\frac{1}{3^{2018}}}{2}\)
đặt \(B=1+5+5^2+...+5^{2018}\)
\(5B=5+5^2+5^3+...+5^{2019}\)
\(5B-B=\left(5+5^2+5^3+...+5^{2019}\right)-\left(1+5+5^2+...+5^{2018}\right)\)
\(4B=5^{2019}-1\)
\(B=\frac{5^{2019}-1}{4}\)
XC
1
TC
22 tháng 9 2019
A = 1+5^2+5^3+5^4+...+5^2018+5^2019
5A = 5^1+5^3+5^4+...+5^2018+5^2019+5^2020
5A - A = 5^2020 + 5 - 1
4A = 5^2020 + 4
4A + 1 = 5^2020 + 4 - 1
4A - 1 = 5^2020 + 3
TN
0
PH
1
NY
3
3 tháng 1 2019
B=1+1/5+1/52+...+1/52018
=>5B=5+1+1/5+...+1/52017
=>5B-B=5-1/52018
=>4B=5-1/52018
=>B=(5-1/52018)/4
3 tháng 1 2019
\(B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\)
\(\Rightarrow5B=5\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\right)\)
\(\Rightarrow5B=5+1+\frac{1}{5}+...+\frac{1}{5^{2017}}\)
\(\Rightarrow5B-B=\left(5+1+\frac{1}{5}+...+\frac{1}{5^{2017}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\right)\)
\(\Rightarrow4B=5-\frac{1}{5^{2018}}\)
\(\Rightarrow B=\frac{5-\frac{1}{5^{2018}}}{4}\)
Vậy \(B=\frac{5-\frac{1}{5^{2018}}}{4}\)
S
25 tháng 3 2023
(Dấu . là dấu nhân)
a/\(\dfrac{2}{5}\cdot\dfrac{4}{3}-\dfrac{2}{5}:3\)
\(=\dfrac{2}{5}\cdot\dfrac{4}{3}-\dfrac{2}{5}\cdot\dfrac{1}{3}\)
\(=\dfrac{2}{5}\cdot\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)
\(=\dfrac{2}{5}\cdot1\)
\(=\dfrac{2}{5}\)
b/\(\dfrac{2010}{2018}:\dfrac{1}{2}+\dfrac{7}{2018}:\dfrac{1}{2}\)
\(=\left(\dfrac{2010}{2018}+\dfrac{7}{2018}\right):\dfrac{1}{2}\)
\(=\dfrac{2017}{2018}:\dfrac{1}{2}\)
\(=\dfrac{2017}{2018}\cdot2\)
\(=\dfrac{2017}{1009}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 3 2023
a, \(\dfrac{2}{5}\) \(\times\) \(\dfrac{4}{3}\) - \(\dfrac{2}{5}\) : 3
= \(\dfrac{2}{5}\) \(\times\) \(\dfrac{4}{3}\) - \(\dfrac{2}{5}\) \(\times\) \(\dfrac{1}{3}\)
= \(\dfrac{2}{5}\) \(\times\) ( \(\dfrac{4}{3}\) - \(\dfrac{1}{3}\))
= \(\dfrac{2}{5}\) \(\times\) 1
= \(\dfrac{2}{5}\)
b, \(\dfrac{2010}{2018}\) : \(\dfrac{1}{2}\) + \(\dfrac{7}{2018}\) : \(\dfrac{1}{2}\) + \(\dfrac{1}{2018}\) : \(\dfrac{1}{2}\)
= \(\dfrac{2010}{2018}\) \(\times\) \(\dfrac{2}{1}\) + \(\dfrac{7}{2018}\) \(\times\) \(\dfrac{2}{1}\) + \(\dfrac{1}{2018}\) \(\times\) \(\dfrac{2}{1}\)
= \(\dfrac{2}{1}\) \(\times\) ( \(\dfrac{2010}{2018}\) + \(\dfrac{7}{2018}\) + \(\dfrac{1}{2018}\))
= 2 \(\times\) \(\dfrac{2018}{2018}\)
= 2 \(\times\) 1
= 2
Theo đề
=>5K=5+52+53+...+52019
=>5K-K=(5+52+53+...+52019)-(1+5+52+...+52018)
=>4K=52019-1
=>K=\(\frac{5^{2019}-1}{4}\)
Chúc bạn học tốt!
K = 1+5+52+...+52018
5K=5+52+...+52018+52019
5K-K=52019-1
4K=52019-1
K=\(\frac{5^{2019}-1}{4}\)
Vì số to quá nên ko tính được ra hết đâu bạn ạ