The sum of series 1+13.14+15.142+17.143+….is
loge1
loge2
loge3
loge4
Explanation for the correct option:
Find the required sum.
Given: series 1+13.14+15.142+17.143+….
We know that, log(1+x)=x–x22+x33–x44+……
and log(1-x)=-x–x22–x33–x44–……
So,
log(1+x)−log(1−x)=x–x22+x33–x44+……--x–x22–x33–x44–……⇒log(1+x)−log(1−x)=x–x22+x33–x44+……+x+x22+x33+x44+……⇒log(1+x)−log(1−x)=2x+2x33+2x55……⇒log(1+x)−log(1−x)=2x+x33+x55……
Put x=12 on both sides, we get
⇒log1+12−log1−12=212+1233+1255+……⇒log32−log12=212+13123+15125+……⇒log3212=1+13122+15124+……∵logm-logn=logmlogn⇒loge3=1+1314+15142+……
Hence, option (C) is the correct answer.
Find the value of x so that; (i) (34)2x+1=((34)3)3(ii) (25)3×(25)6=(25)3x(iii) (−15)20÷(−15)15=(−15)5x(iv) 116×(12)2=(12)3(x−2)