(-10):(n-3)
Bài 1. Chứng minh
a, 10^ 2020 + 10^ 2021 + 10^ 2022 chia hết cho 222
b, 81^ 7 – 27^ 9 – 9^ 13 chia hết cho 45
c, 10^ 6 – 5 ^7 chia hết cho 59
d, 24^ 54 .54^ 24 .2^ 10 chia hết cho 72 ^63
e,3^ n+2 – 2^ n+2 + 3^ n – 2 ^n chia hết cho 10;
f, 3^ n+3 + 3^ n+1 + 2^ n+3 + 2^ n+2 chia hết cho 6
Bài 2.
a, Cho A = 1 + 2 + 2 ^2 + 2 ^3 + ...+ 2^ 99 . Chứng tỏ A chia hết cho 3; A chia 7 dư 1.
b, Cho B = 2 + 2^ 2 + 2^ 3 + ...+ 2^ 99 + 2^ 100 . Hỏi A có chia hết cho 6 không?
Bài 3. Cho A = 9^ 7 + 3^ 13 + 2. Hỏi A có chia hết cho 10 không?
ChoA=(10^n+10^n-1+10^n-2...+10+1)×(10^n+1 +3)+11la so chinh phuong
Cho A=(10^n+10^n-1...+10+1)×(10^n+1 +3)+1la so cp
hãy giải thích tại sao 3^n.10-2^n.5=3^n.10-2^n-1.10
Rút gọn biểu thức:
a) 10^n+1-6*10^n
b) 90*10^n-10^n-2+10^n+1
c) 2,5 *56^n-3
a) \(10^n+1-6\cdot10^n=\left(1-6\right)10^n+1=-5\cdot10^n+1\)
b) \(90\cdot10^n-10^2-2+10^n+1=\left(90-1+1\right)\cdot10^n-2+1=90\cdot10^n-1\)
c) \(2,5\cdot56^n-3=\frac{5}{2}\cdot56^n-3\)
2^10+2^2^10+2^3^10+...+2^10^10=2^n tim n
cho mk hỏi là đây là lũy thừa tầng hay binh thường
(10^3-1)(10^3-2)(10^3-3)...(10^3-n) (tích có 1500 thừa số)
tim x thuoc N bieta, 125.n 5 7b, 2 3.n 3 4 2 5 5c, 2 3 2 n 3 2.n.5 10 10 2d, 5 n=
125
1) Tính:
a) (10 . 102 . 103 .104 ....................109 ) : ( 105 . 1010 . 1025 )
2) Cho A: 2 + 22 + 23 + ......................+ 250 .Chứng tỏ A chia hết cho 3
3) Tìm x e n sao cho : 42: (2x + 3)
4) Cho A = 3 + 32 + 33 + ................. + 32014 Tìm n e N biết 2 A + 3 = 3n
Giải giúp minh những bai nay nhé , minh cảm ơn nhiều lắm!
1) Tính \(S=-1+\dfrac{1}{10}-\dfrac{1}{10^2}+...+\dfrac{\left(-1\right)^n}{10^{n-1}}\)
2) Tính \(S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{n-1}}\)
1:
\(S=-\left(1-\dfrac{1}{10}+\dfrac{1}{10^2}-...-\dfrac{1}{10^{n-1}}\right)\)
\(=-\left[\left(-\dfrac{1}{10}\right)^0+\left(-\dfrac{1}{10}\right)^1+...+\left(-\dfrac{1}{10}\right)^{n-1}\right]\)
\(u_1=\left(-\dfrac{1}{10}\right)^0;q=-\dfrac{1}{10}\)
\(\left(-\dfrac{1}{10}\right)^0+\left(-\dfrac{1}{10}\right)^1+...+\left(-\dfrac{1}{10}\right)^{n-1}\)
\(=\dfrac{\left(-\dfrac{1}{10}\right)^0\left(1-\left(-\dfrac{1}{10}\right)^{n-1}\right)}{-\dfrac{1}{10}-1}\)
\(=\dfrac{1-\left(-\dfrac{1}{10}\right)^{n-1}}{-\dfrac{11}{10}}\)
=>\(S=\dfrac{1-\left(-\dfrac{1}{10}\right)^{n-1}}{\dfrac{11}{10}}\)
2:
\(S=\left(\dfrac{1}{3}\right)^0+\left(\dfrac{1}{3}\right)^1+...+\left(\dfrac{1}{3}\right)^{n-1}\)
\(u_1=1;q=\dfrac{1}{3}\)
\(S_{n-1}=\dfrac{1\cdot\left(1-\left(\dfrac{1}{3}\right)^{n-1}\right)}{1-\dfrac{1}{3}}\)
\(=\dfrac{3}{2}\left(1-\left(\dfrac{1}{3}\right)^{n-1}\right)\)
\(1,\) Ta có \(\left\{{}\begin{matrix}q=\dfrac{u_2}{u_1}=\dfrac{1}{10}:\left(-1\right)=-\dfrac{1}{10}\\u_1=-1\end{matrix}\right.\)
Vậy \(S=-1+\dfrac{1}{10}-\dfrac{1}{10^2}+...+\dfrac{\left(-1\right)^n}{10^{n-1}}=\dfrac{-1}{1-\left(-\dfrac{1}{10}\right)}=-\dfrac{10}{11}\)
\(2,\) Ta có \(\left\{{}\begin{matrix}q=\dfrac{u_2}{u_1}=\dfrac{1}{3}\\u_1=1\end{matrix}\right.\)
Vậy \(S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{n-1}}=\dfrac{1}{1-\dfrac{1}{3}}=\dfrac{3}{2}\)