Expansion frenzy

Correct answer: $2$ .

In order to find x for the above expansion, we must first find n. Knowing that the sum of the coefficients of the last three terms is 22 gives us the following:

${n\choose n}+{n\choose n-1}+{n\choose n-2}=22 \Rightarrow 1+n+\frac{n(n-1)}{2}=22 \Rightarrow 2+2n+n^{2}-n=44 \Rightarrow n^{2}+n-42=0$ . From the quadratic equation we get the solutions $n_{1}=6$ and $n_{2}=-7$ . Since $n\ge 2$ for the binomial expression, we get that $n=6$ .

Now, in order to find x, we must first find the third and the fifth term:

$T_{3}={6\choose 2}(\sqrt{2^{x}})^{4} \times \frac{(\sqrt{2})^{2}}{(\sqrt{2^{x}})^{2}}=15(2^{x})(2)$ and $T_{5}={6\choose 4}(\sqrt{2^{x}})^{2} \times \frac{(\sqrt{2})^{4}}{(\sqrt{2^{x}})^{4}}=15(\frac{4}{2^{x}})$ .

If we add the two terms we get: $T_{3}+T_{5}=135 \Rightarrow 30(2^{x})+\frac{60}{2^{x}}=135 \Rightarrow 2(2^{x})+\frac{4}{2^{x}}=9$ . By multiplying with $2^{x}$ we get: $2(2^{x})^{2}+4=9(2^{x})$ . By replacing $2^{x}=t , t> 0$ , we get the quadratic equation $2t^{2}-9t+4=0$ , which we solve for $t$ , obtaining $t_{1}=4$ and $t_{2}=\frac{1}{2}$ , from which we obtain $x_{1}=2$ and $x_{2}=-1$ . We notice that by replacing both of the solutions in the given expansion, the result does not change, therefore $x=2$ .