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gọi biểu thức trên là A , ta có :
\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+\dfrac{5}{3^5}-...+\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}\\ 3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\ \Rightarrow A+3A=\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)+\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)\\ \Rightarrow4A\cdot3=12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)
từ đó ta được :
\(16A=3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\\ \Rightarrow A=\dfrac{\dfrac{3-101}{3^{99}}-\dfrac{100}{3^{100}}}{16}\\ \Rightarrow A=\dfrac{3}{16}-\dfrac{\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{16}< \dfrac{3}{16}\)
Bài làm:
a) \(a=2+2^3+2^5+...+2^{99}+2^{101}\)
\(\Rightarrow4a=2^3+2^5+2^7+...+2^{101}+2^{103}\)
\(\Rightarrow4a-a=\left(2^3+2^5+2^7+...+2^{103}\right)-\left(2+2^3+2^5+...+2^{101}\right)\)
\(\Leftrightarrow3a=2^{103}-2\)
\(\Rightarrow a=\frac{2^{103}-2}{3}\)
Vậy \(a=\frac{2^{103}-2}{3}\)
b) \(b=1-5^3+5^6-5^9+...+5^{96}-5^{99}\)
\(\Rightarrow125b=5^3-5^6+5^9-5^{12}+...+5^{99}-5^{102}\)
\(\Rightarrow125b+b=\left(5^3-5^6+5^9-5^{12}+...+5^{99}-5^{102}\right)+\left(1-5^3+5^6-5^9+...+5^{96}-5^{99}\right)\)
\(\Leftrightarrow126b=1-5^{102}\)
\(\Rightarrow b=\frac{1-5^{102}}{126}\)
Vậy \(b=\frac{1-5^{102}}{126}\)
Học tốt!!!!
a)64:2mũ5×30×4
= 64 : 32 x 30 x 4
= 240
b)3 mũ 2× 5 - 2 mũ 2×7+2 mũ 0 × 5
= 9 x 5 - 4 x 7 + 1 x 5
= 45 - 28 + 5
= 22
c)2 mũ 3-5 mũ 3÷5 mũ 2 + 12×2 mũ 2
= 8 - 125 : 25 + 12 x 4
= 8 - 5 + 48
= 51
d)2[(7-3 mũ 3÷3 mũ 2) chia 2 mũ 2 + 99]-100
= 2[( 7 - 27 : 9) : 4 + 99] - 100
= 2[4 : 4 + 99] - 100
= 2. 100 - 100
= 200 - 100
= 100
e)4[(3 + 3^7:3^4)chia 10 + 97]-300
= 4[( 3 + 3^3) : 10 + 97] - 300
= 4[ 30 : 10 + 97 ] - 300
= 4. 100 - 300
= 400 - 300
= 100
f)2^2 x 5 [(5 mũ 2 cộng 2 mũ 3) chia 11 - 2] - 3^2 x 2
= 4 x 5 [ (25 + 8 ) : 11 - 2] - 9 x 2
= 20 [ 33 : 11 - 2] - 18
= 20. 1 - 18
= 20 - 18
= 2
\(A=5+5^2+5^3+...+5^{99}+5^{100}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{99}\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{99}.6\)
\(=6.\left(5+5^3+...+5^{99}\right)\)
Vì \(A=6.\left(5+5^3+...+5^{99}\right)\)lên A chia hết cho 6.
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