SAT1000 - P496

Given that $a_n = n^2 \left(\cos^2 \dfrac {n\pi}3 - \sin^2 \dfrac {n\pi}3 \right) = n^2 \cos \dfrac {2n\pi}3 = \begin{cases} - \dfrac {n^2}2 & \text{if } n \bmod 3 \ne 0 \\ n^2 & \text{if } n \bmod 3 = 0 \end{cases}$

Now consider

$\begin{aligned} S_{3n} & = - \frac {1^2}2 - \frac {2^2}2 + 3^2 - \frac {4^2}2 - \frac {5^2}2 + 6^2 - \cdots - \frac {(3n-2)^2}2 - \frac {(3n-1)^2}2 + (3n)^2 \\ & = \frac 32 \sum_{k=1}^n (3k)^2 - \frac 12 \sum_{k=1}^{3n} k^2 = \frac {27}2 \sum_{k=1}^n k^2 - \frac 12 \sum_{k=1}^{3n} k^2 \\ & = \frac {27}2 \cdot \frac {n(n+1)(2n+1)}6 - \frac 12 \cdot \frac {3n(3n+1)(6n+1)}6 \\ & = \frac {n(9(2n^2+3n+1)-(18n^2+9n+1)}4 \\ & = \frac {n(9n+4)}2 \end{aligned}$

Then we have:

$\begin{aligned} \implies S_{3n+1} & = \frac {n(9n+4)}2 - \frac {(3n+1)^2}2 \\ \implies S_{3n-1} & = \frac {n(9n+4)}2 - (3n)^2 \\ S_{100} & = S_{3\cdot 33+1} = \frac {33(9\cdot 33+4)}2 - \frac {100^2}2 = -33.5 \\ S_{201} & = S_{3\cdot 67} = \frac {67(9\cdot 67+4)}2 = 20334.5 \\ S_{302} & = S_{3\cdot 101-1} = \frac {101(9\cdot 101+4)}2 - 303^2 = -45702.5 \end{aligned}$

And:

$\begin{aligned} \implies b_n & = \frac {S_{3n}}{4^n n} = \frac {9n+4}{2 \cdot 4^n} \\ \implies T_n & = \sum_{k=1}^n b_k = \sum_{k=1}^n \frac {9k+4}{2 \cdot 4^k} = \frac 98 \sum_{k=1}^n \frac k{4^{k-1}} + \frac 12 \sum_{k=0}^{n-1} \frac 1{4^k} \\ & = \frac 98 \sum_{k=1}^n \frac d{dx} x^k \bigg|_{x = \frac 14} + \frac 12 \left(\frac {1-\frac 1{4^n}}{1-\frac 14}\right) \\ & = \frac 98 \cdot \frac d{dx} \sum_{k=1}^n x^k \bigg|_{x = \frac 14} + \frac 12 \left(\frac {4^n-1}{4^n-4^{n-1}}\right) \\ & = \frac 98 \cdot \frac d{dx} \left[\frac {x(1-x^n)}{1-x} \right]_{x = \frac 14} + \frac {4^n-1}{6\cdot 4^{n-1}} \\ & = \frac 98 \left[\frac {nx^{n+1}-(n+1)x^n+1}{(1-x)^2} \right]_{x = \frac 14} + \frac {4^n-1}{6\cdot 4^{n-1}} \\ & = 2 \left(\frac {4^{n+1}-3n -4}{4^{n+1}} \right) + \frac {4^n-1}{6\cdot 4^{n-1}} \\ & = \frac {16(4^n-1)-9n}{6\cdot 4^n} \\ \implies T_{10} & = \frac {16777110}{6291456} = \frac {2796185}{1048576} \end{aligned}$

Therefore $\left \lfloor p-q+2(S_{100}+S_{201}+S_{302} \right \rfloor = \boxed{1696806}$ .

$S_{3n}=\dfrac {n}{2}(9n+4)\implies S_{201}=\dfrac{40669}{2}$

$S_{3n+1}=-\dfrac {2n+1}{2}\implies S_{100}=-\dfrac {67}{2}$

$S_{3n+2}=-\dfrac {9n^2+14n+5}{2}\implies S_{302}=-\dfrac {91405}{2}$

$b_n=\dfrac {9n+4}{2^{2n+1}}\implies$

$T_n=\dfrac {2^{2n}-1}{3\times 2^{2n-3}}-\dfrac {3n}{2^{2n+1}}$

$\implies T_{10}=\dfrac {2796185}{1048576}$

Hence the required answer is $\boxed {1696806}$ .