Even perfect squares; 4, 16, 36, 64, 100, ...
Odd perfect cubes: 1, 27, 125...
Not enough information, but use that.
(x^4 + x^3 + x^2 + x + 1) + (x^4 - x^3 + x^2 - x + 1)
\(2x^4 + 2x^2 + 1\)
So far I get \(a_1 = 5\), \(a_2 = 10\), and \(a_3 = 20\), getting \(35\). Not sure.
\(1.11111 + 0.11111 + 0.01111 + 0.00111 + 0.00011 + 0.00001 = 1.23456\)
it comes from the MathCounts 2022 National Sprint. Question #23.
\(\frac{1}{\sqrt{36}} - \sqrt{27} - \frac{1}{\sqrt{27}} - \sqrt{18} + \frac{1}{\sqrt{18}} - \sqrt{9}\)
is equal to
\(-\frac{17}{6} + \frac{1}{3\sqrt2} - 3\sqrt2 - \frac{1}{3\sqrt{3}} - 3\sqrt{3}\)
rationalizing the denominators and simplifying gives\(-\frac{17}{6} + \frac{\sqrt2}{6} - 3\sqrt2 - \frac{\sqrt{3}}{9} - 3\sqrt{3}\)
\(\text{WolframAlpha also works.}\)
\(\text{In the end, I was able to find the expansion of}\)
\({(x + y)}^{31}\) correctly.
Unfortunately, it doesn't show any steps (or minimal)
Using brute force , which is something you should NOT do (even though this time, I'm just lazy):
\(11250x^6 + 12750(x^5)(y) - 1800(x^4)(y^2) - 4740(xy)^3\)\( - 1662(x^2)(y^4)-234(x)(y^5) - 12y^6\)
\(\text{The coefficient of} (x^2)(y^4) \text{ is -1,662.}\)
Hope this helped .
+51 
, which is something you should NOT do (even though this time, I'm just lazy):