Tính giá trị của biểu thức A=log (1+1/1) + log (1+1/2)+ log (1+1/3)+...+ log(1+1/99)

Trả lời

$A=\log \left(1+\frac{1}{1}\right)+log\left(1+\frac{1}{2}\right)+log\left(1+\frac{1}{3}\right)+.\text{\hspace{0.33em}}..+log\left(1+\frac{1}{99}\right)$

$=\log 2+\log \frac{3}{2}+\log \frac{4}{3}+....+\log \frac{100}{9}$

$=\log \left(2\cdot \frac{3}{2}\cdot \frac{4}{3}\cdot ....\cdot \frac{100}{9}\right)=\log 100=2$