$\displaystyle T_{n}=4T_{n-1}-3n, T_{0}=5$
Find $\displaystyle \sum_{3}^{12}T_{n}$
Is there a way to find this summation without listing out all the terms?
Thanks!
$\displaystyle T_n=\frac{1}{3} \left(4+11\cdot 4^n+3 n\right)$
$\displaystyle \sum _{n=3}^{12}\frac{1}{3} \left(4+11\cdot 4^n+3 n\right)=82021955$
I used Wolfram Mathematica. I hope I didn't make any mistakes. (I don't know how can it be done with pure maths.)
$\displaystyle \sum_{n=3}^{12} T_n = \sum_{n=3}^{12} 4T_{n-1} - 3n = \sum_{n=2}^{11}4T_n - 3(n+1) $
Thus,
$\displaystyle \sum_{n=3}^{12}T_n = 4\left( \sum_{n=3}^{12} T_n \right) - \left( 3\sum_{n=3}^{12} n \right) - 4(T_3 + T_{12})$
Therefore,
$\displaystyle -3\left( \sum_{n=3}^{12}T_n \right) = -3 \left( \frac{(12)(13)}{2} - \frac{(3)(4)}{2} \right) - 4(T_{12} + T_3)$
It remains to find $\displaystyle T_{12}$ and $\displaystyle T_3$.
This should not be hard since you can skip successive steps.