Find the sum of this series \displaystyle \log\left(\frac{1}{2}\right)+\log\l eft(\frac{2}{3}\right)+\log\left(\f rac{3}{4}\right)+\log\left(\frac{4} {5}\right)+...+\log\left(\frac{98}{ 99}\right) +\log\left(\frac{99}{100}\right)log ( 2 1 ​ )+log( 3 2 ​ )+log( 4 3 ​ )+log( 5 4 ​ )+...+log( 99 98 ​ )+log( 100 99 ​ )

Step-by-step explanation:

Given

Required

The sum

Using the laws of logarithm, we have:

Take the first two terms of the series

Include the third term

Include the fourth term

Notice the following pattern

---------------- n =2

-------------- n = 3

----------- n = 4

So the sum of n series is:

So, the sum of the series is:

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11