If|a|<1,b=∑n=1∞akk then a is equal to
(1)∑k=1∞(-1)kbkk
(1)∑k=1∞(-1)k-1bkk!
(1)∑k=1∞(-1)k-1bk(k-1)!
None of these
Explanation for the correct option:
Find the value of a:
Given |a|<1,b=∑n=1∞akk
=a11+a22+a33+...........∞=-log(1-a)[∵log(1-x)=-(x+x22+x33+........∞)]
⇒e-b=(1-a)⇒a=1-e-b⇒a=1-1-b+b22!-b33!+........∞⇒a=b-b22!+b33!+........∞⇒a=∑k=1∞(-1)k-1bkk!
Hence, the correct option is B.