$$(1+5)\log_{e}3+\frac{(1+5^{2})}{2!}(\log_{e}3)^{2}+\frac{(1+5^{3})}{3!}(\log_{e}3)^{3}+...$$

Splitting the above mentioned series into two series 

A = $$\log_{e}3+\frac{1}{2!}(\log_{e}3)^{2}+\frac{1}{3!}(\log_{e}3)^{3}+...$$

B = $$5\log_{e}3+\frac{5^{2}}{2!}(\log_{e}3)^{2}+\frac{5^{3}}{3!}(\log_{e}3)^{3}+...$$

We know that $$e^{x}$$ =$$1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$$

So  $$e^{x}-1$$ = $$x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$$

On solving two series A and B

A = $$\log_{e}3+\frac{1}{2!}(\log_{e}3)^{2}+\frac{1}{3!}(\log_{e}3)^{3}+...$$ =$$e^{\log_{e}3}-1$$ = $$3-1$$ =$$2$$

B = $$5\log_{e}3+\frac{5^{2}}{2!}(\log_{e}3)^{2}+\frac{5^{3}}{3!}(\log_{e}3)^{3}+...$$=$$e^{\log_{e}3^{5}}-1$$=$$3^{5}-1$$=$$242$$

A+B = $$2 + 242$$ = $$244$$