Đặt : \(A=5+5^2+5^3+...+5^{30}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{29}\left(1+5\right)\)

\(=\left(1+5\right)\left(5+5^3+...+5^{29}\right)\)

\(=6\left(5+5^3+...+5^{29}\right)⋮6\) (đpcm)

                                                   Bài giải

\(5+5^2+5^3+5^4+...+5^{29}+5^{30}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{29}\left(1+5\right)\)

\(=5\cdot6+5^3\cdot6+...+5^{29}\cdot6\)

\(=6\left(5+5^3+...+5^{29}\right)\text{ }⋮\text{ }6\)

\(\Rightarrow\text{ ĐPCM}\)

Bài 3:

\(A=5+5^2+..+5^{12}\)

\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)

\(5A=5^2+5^3+...+5^{13}\)

\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)

\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)

\(4A=5^{13}-5\)

\(A=\dfrac{5^{13}-5}{4}\)

a)Ta có:

6¹⁰⁰-1=...6-1=...5 chia hết cho 5

=>6¹⁰⁰-1 chia hết cho 5(đpcm)

b)Ta có:

21²⁰-11¹⁰=...1-...1=...0 chia hết cho 5

=>21²⁰-11¹⁰ chia hết cho 5(đpcm)

CÂU 1 TÍNH RA KHẮC BIẾT

CÂU 2 TA CÓ:    X/3 + X/4 = 21/ 12

                         X.(1/3 + 1/4) = 7/4

                         X. 7/12      = 7/4

                           X = 3

S = (1+5)+(5^2+5^3)+(5^4+5^5)+(5^6+5^7)

   = 6+5^2.(1+5)+5^4.(1+5)+5^6.(1+5)

   = 6+5^2.6+5^4.6+5^6.6

   = 6.(1+5^2+5^4+5^6) chia hết cho 6

=> ĐPCM

k mk nha

(1+5)+(5^2+5^3)+........+(5^6+5^7)

=6+5^2(1+5)+......+5^6(1+5)

=6+5^2 . 6 +.....+5^6 . 6

= 6 ( 5^2+.....+5^6)

Suy ra S chia hết cho 6

Ta có: \(\frac{1}{21}<\frac{1}{20};\frac{1}{22}<\frac{1}{20};\ldots;\frac{1}{40}<\frac{1}{20}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{40}<\frac{1}{20}+\frac{1}{20}+\cdots+\frac{1}{20}=\frac{20}{20}\) =1(1)

Ta có: \(\frac{1}{41}<\frac{1}{40};\frac{1}{42}<\frac{1}{40};\ldots;\frac{1}{60}<\frac{1}{40}\)

Do đó: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{60}<\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{20}{40}=\frac12\) (2)

Từ (1),(2) suy ra \(\left(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{60}\right)\) <1+1/2

=>A<3/2(3)

Ta có: \(\frac{1}{21}>\frac{1}{40};\frac{1}{22}>\frac{1}{40};\ldots;\frac{1}{40}=\frac{1}{40}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{20}{40}\) =1/2(3)

Ta có: \(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};\ldots;\frac{1}{60}=\frac{1}{60}\)

Do đó: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+\cdots+\frac{1}{60}=\frac{20}{60}=\frac13\) (4)

Từ (3),(4) suy ra A>1/2+1/3

=>A>5/6(4)

Từ (3),(4) suy ra 5/6<A<3/2

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