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As the title says, I need to find the explicit form of the recursive sequence defined above, and I am very stuck on this.

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closed as off-topic by John Omielan, pi66, Claude Leibovici, Leucippus, RRL Apr 19 at 7:06

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  • $\begingroup$ Is this the correct formula? $\endgroup$ – amsmath Apr 19 at 2:41
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    $\begingroup$ What have you tried? Have you tried computing the first dozen terms in a spreadsheet? $\endgroup$ – Ross Millikan Apr 19 at 2:42
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So you have $b_k = 2\cdot 3^k + 2\cdot 3^{k-1}+\dots+2\cdot 3^2 + 2$. This is the same as $2\cdot (3^k+3^{k-1}+\dots+3^0) - 2\cdot 3$.

The term in parentheses is a geometric progression. Can you use the geometric progression formula to find a closed form?

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Hint: if you make $b_1=5$ and divide by $2$ you have a geometric series.

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Consider the difference between consecutive terms and take their sum:

$b_k - b_{k-1} = 2 \times 3^k$

Summing over the left hand side gives:

$\sum_{k = 2}^n (b_k - b_{k-1}) = (b_2 - b_1) + (b_3 - b_2) + ... + (b_n - b_{n-1}) = b_n - b_1$

The sum of the RHS from $k = 2$ to $k = n$ can be evaluated using the formula for the sum of terms in a geometric series.

Equate both sums, then solve for $b_k$.

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