Question

$(1+3){\log }_{e}3+\frac{(1+3^2)}{2!}(\log 3{)}^2+\dots ..\infty =$

• A. $27$
• B. $28$
• C. $32$
• D. $33$

Answer 1

Answer: (B)

$28$

$=\log e^3+3\log e^3+\frac{(\log e^3{)}^2}{2!}+\frac{3^2(\log e^3{)}^2}{2!}\cdots$
$=e^{\log e^3}+e^{3\log e^3}-1-1$ $(e^x=1+x+\frac{x^2}{2!}\cdots )$
$=3+27-2$
$=28$