Stewart's Sequences 3 of 5

Firstly, we have: $3x^2 -7x+5=31111$ Rearranging to = 0 gives: $3x^2-7x-31106=0$ Using the quadratic formula to solve the equation: $x=\frac { -b\pm \sqrt { b^{ 2 }-4ac } }{ 2a }$ Inputting our values of $a=3, b=-7, c=-31106$ gives us: $x=\frac { 7\pm \sqrt { {(-7)}^{ 2 }-4\times 3\times -31106 } }{ 2\times 3 }$ Simplifying: $x=\frac { 7\pm \sqrt { 49+373272 } }{ 6 }$ Summing the square root: $x=\frac {7\pm \sqrt {373321}}{6}$ This results in two answers: $x=\frac{7+611}{6}\quad or\quad \frac{7-611}{6}$ $x=103\quad or\quad -\frac{302}{3}$ Now, we're told that $x$ is a positive , non-zero integer . This means that $x=103$ Now, inputting our value for $x$ into the given equation, results in: $103(2\times 103 + 4) + 17$ Which simplifies to: $103\times 210 +17$ Which equals: $21647$ Now, the four options given were: $\begin{matrix} 2^{ 15 }-3^{ 9 }+4^{ 7 }-5^{ 6 }+6^{ 5 } \\ 2^{ 15 }-3^{ 8 }-4^{ 6 }-5^{ 3 }-6^{ 3 }-7^{ 2 }-8^{ 2 }-9^{ 1 }-10^{ 0 } \\ 2^{ 15 }-3^{ 8 }-4^{ 6 }-5^{ 3 }-6^{ 4 }+7^{ 3 }+8^{ 3 }+9^{ 2 }+10^{ 0 } \\ 2^{ 15 }-3^{ 10 }+4^{ 8 }-5^{ 7 }+6^{ 6 }+7^{ 5 }-8^{ 4 }+9^{ 3 } \end{matrix}$ These can be inputted into a calculator, or solved on paper to give: $\begin{matrix} 21620 \\ 21647 \\ 21627 \\ 21226 \end{matrix}$ Therefore, the answer we are looking for is: $\boxed {2^{ 15 }-3^{ 8 }-4^{ 6 }-5^{ 3 }-6^{ 3 }-7^{ 2 }-8^{ 2 }-9^{ 1 }-10^{ 0 }}$ Which gives an answer of $\boxed {21647}$