so sánh 41008/52016và 16504. 32016/52016. 41008
Cho x1−12017=x2−22016=x3−32016=...=x2017−20171x1−12017=x2−22016=x3−32016=...=x2017−20171 và x1+x2+...+x2017=2017.2018
Tìm x1,x2,...,x2017?
Cho x1−12017=x2−22016=x3−32016=...=x2017−20171x1−12017=x2−22016=x3−32016=...=x2017−20171 và x1+x2+...+x2017=2017.2018
Tìm x1,x2,...,x2017?
Cho P=5+52+53+...+52016.Chứng minh P ⋮ 7;9
Tính tổng sau
B = 1 + 31 + 32 + ... + 32016
\(B=1+3^1+3^2+...+3^{2016}\)
\(3B=3+3^2+3^3+3^4+...+3^{2017}\)
\(3B-B=3^{2017}-1\)
\(B=\dfrac{3^{2017}-1}{2}\)
Tính tổng sau
B = 1 + 31 + 32 + ... + 32016
\(B=1+3^1+3^2+...+3^{2016}\)
\(3\cdot B=3+3^2+3^3+...+3^{2016}+3^{2017}\)
\(3B-B=3+3^2+3^3+...+3^{2016}+3^{2017}-\left(1+3^1+3^2+...+3^{2016}\right)\)
\(2B=3^{2017}-1\)
\(\Rightarrow B=\dfrac{3^{2017}-1}{2}\)
Tính tổng sau
B = 1 + 3 + 32 + ... + 32016
\(B=1+3+3^2+...+3^{2016}\)
\(3\cdot B=3+3^2+3^3+...+3^{2017}\)
\(3B-B=3+3^2+3^3+...+3^{2017}-\left(1+3+3^2+...+3^{2016}\right)\)
\(2B=3^{2017}-1\)
\(\Rightarrow B=\dfrac{3^{2017}-1}{2}\)
cho S = 5 + 52 + 53 + 54 + 55 + 56 +...+ 52016. chứng tỏ rằng S chia hết cho 65
mn giúp mk nhé!!
a) Cho A = 1 + 3 + 32 + 33 + ... + 32016 . Tìm số dư khi chia A cho 65 .
Giúp em với ạ
Lời giải:
$A=1+(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2014}+3^{2015}+3^{2016})$
$=1+3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2014}(1+3+3^2)$
$=1+3.13+3^4.13+....+3^{2014}.13$
$=1+13(3+3^4+...+3^{2014})$
$\Rightarrow A-1\vdots 13(1)$
Mặt khác:
$A=1+(3+3^2+3^3+3^4)+....+(3^{2013}+3^{2014}+3^{2015}+3^{2016})$
$=1+3(1+3+3^2+3^3)+....+3^{2013}(1+3+3^2+3^3)$
$=1+(3+...+3^{2013})(1+3+3^2+3^3)$
$=1+40(3+....+3^{2013})$
$\Rightarrow A-1\vdots 5(2)$
Từ $(1); (2)$ mà $(5,13)=1$ nên $A-1\vdots (5.13)$ hay $A-1\vdots 65$
$\Rightarrow A$ chia $65$ dư $1$
Chứng minh rằng S= 3+32 +33 +...+32016 chia hết cho 12, 39
S = 3+3^2 + 3^3 +...+ 3^2016
= (3+3^2+3^3) +...+(3^2014+3^2015+3^2016)
=3(1+3+3^2) +.....+3^2014(1+3+32)
=13 ( 3+...+3^2014 ) chia hết cho 13
cho A = 3+3^2+3^3+.....+32016
Tim n thuoc N sao cho 2.A +3 = 3^n-1
\(A=3+3^2+3^3+...+3^{2016}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2017}\)
\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2016}\right)\)
\(\Rightarrow2A=3^{2017}-3\)
Ta có : \(2A+3=3^n-1\Rightarrow3^{2017}-3+3=3^n-1\)
\(\Rightarrow3^{2017}=3^{n-1}\Rightarrow n-1=2017\Rightarrow n=2018\)
Vậy : n = 2018