Question
Question: \[1 + \frac{\log_{e}x}{1!} + \frac{(\log_{e}x)^{2}}{2!} + \frac{(\log_{e}x)^{3}}{3!} + .....\infty =...
1+1!logex+2!(logex)2+3!(logex)3+.....∞=
A
logex
B
x
C
x−1
D
−loge(1+x)
Answer
x
Explanation
Solution
(0.5)−2(0.5)2+3(0.5)3−4(0.5)4+....
loge23.