Question

tính D = 4.5\(^{100}\) .(\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.............+\dfrac{1}{5^{100}}\) )+1

Answer 1

Đặt N=\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+......+\dfrac{1}{5^{100}}\)

5N=\(1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+..........+\dfrac{1}{5^{99}}\)

5N-N= \(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+.............+\dfrac{1}{5^{99}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+..........+\dfrac{1}{5^{100}}\right)\)

4N=1-\(\dfrac{1}{5^{100}}\) =\(\dfrac{5^{100}-1}{5^{100}}\)

N=\(\dfrac{5^{100}-1}{4.5^{100}}\)

Thay N vào D ,ta có

D= 4.5\(^{100}\).(\(\dfrac{5^{100}-1}{4.5^{100}}\) )+1

D=5\(^{100}\)

Vậy D =5\(^{100}\)