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\(A=2^1+2^2+2^3+2^4+2^5+2^6+2^7+...+2^{99}\)
\(=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+...+\left(2^{97}+2^{98}+2^{99}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+2^7\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)\)
\(=2.7+2^4.7+2^7.7+...+2^{97}.7\)
\(=\left(2+2^4+2^7+...+2^{97}\right).7⋮7\)
\(\Rightarrow A⋮7\)
A = 21 +22 +23 +24 +25 +26 +27 ….+ 299
A = (21 +22 +23) +(24 +25 +26) + ….+ (297+298+299)
A = 14 + (21.23 +22.23 +23.23) + ….+ (21.296+22.296+23.296)
A = 14 + 23(21+22+23) + ...... + 296(21+22+23)
A = 14.1 + 23.14 + ....... + 296.14
A = 14.(1+23+....+296)
14 \(⋮\) 7
=> A \(⋮\) 7 (đpcm)
a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)
c: \(5S_3=5^6+5^7+...+5^{101}\)
\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)
hay \(S_3=\dfrac{5^{101}-5^5}{4}\)
d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
A=1+3+32+33+...+320
A=(1+3)+(32+33)+(34+35)+...+(319+320)
A= 4+32(1+3)+34(1+3)+......+319(1+3)
A=4+32.4+34.4+....+319.4
A=4.(32+34+...+319) =>A chia hết cho 4
0+(
\(7^{2x-3}=7^2\cdot5+7^2\cdot2\)
\(7^{2x-3}=7^2\left(5+2\right)\)
\(7^{2x-3}=7^2\cdot7\)
\(7^{2x-3}=7^3\)
\(2x-3=3\)
\(2x=6\)
\(x=3\)
\(7^{2x-3}-2.49=49.5\)
\(7^{2x-3}-2.49=245\)
\(7^{2x-3}-2=245:49\)
\(7^{2x-3}-2=5\)
\(7^{2x-3}=5+2\)
\(7^{2x-3}=7\)
2x-3=1
2x=1+3
2x=4
x=4:2
x=2
vậy x=2
S 1 = 1 + 2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7
S 1 = ( 1 + 2 + 2 2 + 2 3 ) + ( 2 4 + 2 5 + 2 6 + 2 7 )
S 1 = ( 1 + 2 + 2 2 + 2 3 ) + ( 1 + 2 + 2 2 + 2 3 ) . 2 4
S 1 = 15 + 15 . 2 4
S 1 = 15 ( 1 + 2 4 )
Vì 15 chia hết cho 5
=> S 1 = 15 ( 1 + 2 4 ) chia hết cho 5
Vậy S 1 chia hết cho 5