Question

a, T= (1-\(\dfrac{1}{2}\))+(1-\(\dfrac{1}{4}\))+(1-\(\dfrac{1}{8}\))+...+(1-\(\dfrac{1}{512}\))+(1-\(\dfrac{1}{1024}\))+(1-\(\dfrac{1}{2048}\))+(1-\(\dfrac{1}{4096}\))

b, 4*5¹⁰⁰*(\(\dfrac{1}{5}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{5^3}\)+...+\(\dfrac{1}{5^{100}}\))+1

Answer 1

\(a,\\ T=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{8}\right)+...+\left(1-\dfrac{1}{4096}\right)\\ T=\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{4096}\right)\)

Gọi \(D=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{4096}\)

\(2D=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2048}\\ 2D-D=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2048}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{4096}\right)\\ D=1-\dfrac{1}{4096}\)

(mk nhớ có cách khác rất hay nhưng quên mất rồi)

Thay \(D\) vào ta được

\(T=\left(1+1+1+...+1\right)-\left(1-\dfrac{1}{4096}\right)\\ T=12-\left(1-\dfrac{1}{4096}\right)\\ T=12-1+\dfrac{1}{4096}\\ T=11\dfrac{1}{4096}\)