# Explicit form of $b_1= 2, b_k = b_{k-1} + 2\cdot 3^k$ for all integers $k\ge 2$ [closed]

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

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

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• Is this the correct formula? – amsmath Apr 19 at 2:41
• What have you tried? Have you tried computing the first dozen terms in a spreadsheet? – Ross Millikan Apr 19 at 2:42

## 3 Answers

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?

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

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 $$b_k$$.