Question

hepl me

\(\dfrac{1+3^4+3^8+3^{12}}{1+3^2+3^4+3^6+3^83+3^{10}+3^{12}+3^{14}}\)

Answer 1

\(\dfrac{1+3^4+3^8+3^{12}}{1+3^2+3^4+3^6+3^8+3^{10}+3^{12}+3^{14}}\)

\(=\dfrac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+\left(3^2+3^6+3^{10}+3^{14}\right)}\)

\(=\dfrac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+3^2\left(1+3^4+3^8+3^{12}\right)}\)

=\(\dfrac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)\left(1+3^2\right)}\)

=\(\dfrac{1}{1+3^2}=\dfrac{1}{10}\)

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