Question

`(1+3)log_e3+(1+3^2)/(2!)(log_e3)^2+(1+3^3)/(3!)(log_e 3)^3+....oo=` (a)28 (b) 30 (c)25 (d) 0
A. 18
B. 28
C. 36
D. none of these

Answer 1

Answer:
We have
`(1+3)(log_(e)3)+(1+3^(2))/(2!)(log_(e)3)^(2)+(1+3^(2))/(3!)(log_(e)3)^(3)+..infty`
`={log_(e)3+(log_(e)3))^(2)/(2!)+(log_(e)3)^(3)/(3!)+…}`
`+{(3log_(e)3)+(3 log_(e) 3)^(2)/(2!)+(3 log_(e) 3)^(3)/(3!)+…}`
`=(e^(log_(e)3-1)+(e^(3)log_(e)3-1)=(3log_(e)3)^(2)/(2!)+(3log_(e)3)^(3)/(3!)+..}`
`=(e^(log_(e)3-1))+(e^(3log_(e)-1)=(3-1)+(3^(3)-1)=28`