Question

[mr. killer]

tinh cac tong sau:

a,3/1.6+3/6.11+...+3/496.501

b,1/2³+1/2⁴+...+1/2²⁰¹⁸

Answer 1

[bảo nam trần]

a, Đặt \(A=\dfrac{3}{1.6}+\dfrac{3}{6.11}+...+\dfrac{3}{496.501}\)

\(5A=\dfrac{3.5}{1.6}+\dfrac{3.5}{6.11}+...+\dfrac{3.5}{496.501}\)

\(5A=3\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{496.501}\right)\)

\(5A=3\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)

\(5A=3\left(1-\dfrac{1}{501}\right)\)

\(5A=3\cdot\dfrac{500}{501}\)

\(A=\dfrac{1500}{501}:5\)

\(A=\dfrac{100}{167}\)

b, Đặt \(B=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2018}}\)

\(2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2017}}\)

\(2B-B=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2017}}\right)-\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2018}}\right)\)

\(B=\dfrac{1}{2^2}-\dfrac{1}{2^{2018}}\)