recursive formula

**acc100jt** · October 16th 2008, 08:01 AM

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

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

Thanks!

**james_bond** · October 16th 2008, 08:13 AM

$T_n=\frac{1}{3} \left(4+11\cdot 4^n+3 n\right)$
$\sum _{n=3}^{12}\frac{1}{3} \left(4+11\cdot 4^n+3 n\right)=82021955$
I used Wolfram Mathematica. I hope I didn't make any mistakes. (I don't know how can it be done with pure maths.)

**ThePerfectHacker** · October 16th 2008, 09:14 AM

Originally Posted by acc100jt

$T_{n}=4T_{n-1}-3n, T_{0}=5$
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.

**acc100jt** · October 18th 2008, 08:55 PM

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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