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Math Help - recursive formula

  1. #1
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    recursive formula

    T_{n}=4T_{n-1}-3n, T_{0}=5
    Find \sum_{3}^{12}T_{n}

    Is there a way to find this summation without listing out all the terms?

    Thanks!
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  2. #2
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    T_n=\frac{1}{3} \left(4+11\cdot 4^n+3 n\right)
    \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.)
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  3. #3
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    Quote Originally Posted by acc100jt View Post
    T_{n}=4T_{n-1}-3n, T_{0}=5
    Find \sum_{3}^{12}T_{n}

    Is there a way to find this summation without listing out all the terms?

    Thanks!
    \sum_{n=3}^{12} T_n = \sum_{n=3}^{12} 4T_{n-1} - 3n = \sum_{n=2}^{11}4T_n - 3(n+1)

    Thus,
    \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,
     -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 T_{12} and T_3.
    This should not be hard since you can skip successive steps.
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  4. #4
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    Thanks, ThePerfectHacker

    I tried what u did.

    Correctme if I m wrong. The only way to find T_{12} is by finding T_{11},
    and find T_{11} by finding T_{10} and so on...

    Am I right?
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