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Given that $T_0 = 6$ and that $T_n$ satisfies the recurrence relation

$$T_{n+1} = 20T_n - 8^n \times 48$$

I have the equation for any term $n$ to be;

$T_n = 2\times 20^n+4⋅8^n$

I want to prove this by mathematical induction, however I am unsure as to how to get the equation for the sum of the terms.

This is following on from finding this equation in a previous thread, here. I'd appreciate any help on getting started with this.

Community
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Adrian Hindes
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2 Answers2

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The Base Case: Clearly $2\times 20^0+4⋅8^0 = 6$.

The Inductive Step: Let $n$ be an arbitrary nonnegative integer. Assume that
$$T_n = 2\times 20^n+4⋅8^n.$$

Then \begin{align*} T_{n+1} & = 20\, T_n - 8^n \times 48 \\ & = 20 \bigl(2\times 20^n+4\times8^n\bigr) - 8^n \times 48 \\ & = 2\times 20^{n+1}+80\times8^n - 8^n \times 48 \\ & = 2\times 20^{n+1}+32\times8^n \\ & = 2\times 20^{n+1}+4\times8^{n+1} \\ \end{align*} Thus $$ T_{n+1} = 2\times 20^{n+1}+4\times8^{n+1}. $$

wdacda
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If you just want to prove the statement by M.I., the induction step is just \begin{align} T_{n+1} & = 20T_n - 48 \times 8^n\\ & = 20(2 \times 20^n + 4 \times 8^n) - 48 \times 8^n\\ & = 2 \times 20^{n+1} + (80-48) \times 8^n\\ & = 2 \times 20^{n+1} + 32 \times 8^n\\ & = 2 \times 20^{n+1} + 4 \times 8^{n+1} \end{align}

Empiricist
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  • Oh cool thanks, so I didn't need the equation for the sum of the terms. – Adrian Hindes Jun 03 '15 at 03:46
  • That equation from Andrews is just for you know observing the pattern, but that is essential: imagine if you cannot observe the pattern, you don't even have a "statement" to prove my M.I. – Empiricist Jun 03 '15 at 03:52
  • I was a bit confused as to how ∑(k=0 to n−1) 20^n−1−k 8^k got to 20^n−8^n/20−8 in Andrew's working, although I'm not sure if that warrants a separate question. – Adrian Hindes Jun 03 '15 at 04:05