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The sum of the series "log"(4)2-"log"...

The sum of the series
`"log"_(4)2-"log"_(8)2 + "log"_(16)2- "log"_(32) 2+….,` is

A

`e^(2)`

B

`"log"_(e) 2+1`

C

`"log"_(e)3-2`

D

`1-"log"_(e)2`

Text Solution

Verified by Experts

The correct Answer is:
D
We have,
`"log"_(4) 2-"log"_(8)2+ "log"_(16)2-"log"_(32)+….`
` = (1)/(2) "log"_(2) 2 -(1)/(3) "log"_(2)2 + (1)/(4)"log"_(2)2 - (1)/(5)"log"_(2)2 + ….`
`= (1)/(2) - (1)/(3) + (1)/(4) - (1)/(5) + …. = -"log"_(e)2+1`
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