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Explanation:

Consider log_e3 = x.

Hence, the series can also be written as

$\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\right)+\left(5x+\frac{{\left(5x\right)}^2}{2!}+\frac{{\left(5x\right)}^3}{3!}+...\right)$

Now, each bracket is a standard expression that can be expressed as a power of e. Hence, the expression becomes:

$(e^x-1)+(e^{5x}-1)=e^{{\log }_{e}3}-1+e^{5{\log }_{e}3}-1$

= 3 – 1 + 3⁵ – 1 = 1 + 243 = 244

Hence, option (b).

Note:

​​​​​​​This is one of the rare questions where pure engineering maths based concepts have been used. In the exam, you are advised to leave such a question if you are not conversant with such series.