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a) \(A=7^{13}+7^{14}+7^{15}+7^{16}+...+7^{100}\)
\(A=\left(7^{13}+7^{14}\right)+\left(7^{15}+7^{16}\right)+...+\left(7^{99}+7^{100}\right)\)
\(A=7^{13}\left(1+7\right)+7^{15}\left(1+7\right)+...+7^{99}\left(1+7\right)\)
\(A=7^{13}.8+7^{15}.8+...+7^{99}.8\)
\(A=8.\left(7^{13}+7^{15}+...+7^{99}\right)\)
⇒ \(A⋮8\)
Vậy A chia hết cho 8 (đpcm)
a) A = 7¹³ + 7¹⁴ + 7¹⁵ + 7¹⁶ + ... + 7⁹⁹ + 7¹⁰⁰
= (7¹³ + 7¹⁴) + (7¹⁵ + 7¹⁶) + ... + (7⁹⁹ + 7¹⁰⁰)
= 7¹³.(1 + 7) + 7¹⁵.(1 + 7) + ... + 7⁹⁹.(1 + 7)
= 7¹³.8 + 7¹⁵.8 + ... + 7⁹⁹.8
= 8.(7¹³ + 7¹⁵ + ... + 7⁹⁹) ⋮ 8
Vậy A ⋮ 8
b) B = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰⁰
= 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + 2⁸ + ... + 2¹⁹⁷ + 2¹⁹⁸ + 2¹⁹⁹ + 2²⁰⁰
= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁹⁷ + 2¹⁹⁸ + 2¹⁹⁹ + 2²⁰⁰)
= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + 2¹⁹⁶.(2 + 2² + 2³ + 2⁴)
= 30 + 2⁴.30 + ... + 2¹⁹⁶.30
= 30.(1 + 2⁴ + ... + 2⁹⁶)
= 5.6.(1 + 2⁴ + ... + 2¹⁹⁶) ⋮ 5
Vậy B ⋮ 5
1.
Ta có:
aaabbb= aaa000+bbb
= a . 111000 + b .111
Vì 111000 \(⋮\) 111 => a.111000 \(⋮\) 111 (1)
111 \(⋮\) 111 => b.111 \(⋮\) 111 (2)ư
Từ (1) và (2) => a.111000 + b.111 \(⋮\) 111
=> aaabbb \(⋮\) 111 (đpcm)
\(a,S=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{19}+3^{20}\right)\\ S=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{18}\left(3+3^2\right)\\ S=\left(3+3^2\right)\left(1+3^2+...+3^{18}\right)=12\left(1+3^2+...+3^{18}\right)⋮12\)
\(b,S=\left(3+3^2+3^3+3^4\right)+...+\left(3^{17}+3^{18}+3^{19}+3^{20}\right)\\ S=\left(3+3^2+3^3+3^4\right)+....+3^{16}\left(3+3^2+3^3+3^4\right)\\ S=\left(3+3^2+3^3+3^4\right)\left(1+...+3^{16}\right)\\ S=120\left(1+...+3^{16}\right)⋮120\)
\(a,S=3+3^2+3^3+...+3^{20}\)
Ta thấy:\(3+3^2=12⋮12\)
\(\Rightarrow S=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{18}\left(3+3^2\right)\\ \Rightarrow S=\left(3+3^2\right)\left(1+3^2+...+1^{18}\right)\\ \Rightarrow S=12.\left(1+3^2+...+3^{18}\right)⋮12\\ \left(đpcm\right)\)
\(b,Ta\) \(thấy:\)\(3+3^2+3^3+3^4=120⋮120\)
\(\Rightarrow S=\left(3+3^2+3^3+3^4\right)+...+\left(3^{17}+3^{18}+3^{19}+3^{20}\right)\\ \Rightarrow S=\left(3+3^2+3^3+3^4\right)+...+3^{16}\left(3+3^2+3^3+3^4\right)\\ \Rightarrow S=\left(3+3^2+3^3+3^4\right)\left(1+...+3^{16}\right)\\ \Rightarrow S=120\left(1+...+3^{16}\right)⋮120\\ \left(đpcm\right)\)
2S=2^1+2^2+...+2^203+2^204
S=2^0+2^1+...+2^202+2^203
=>2S-S=S=2^204-1