5710 Shares

# Use iteration to guess an explicit formula for the sequence?

Topic: Types of parenthesis in math
July 19, 2019 / By Chonsie
Question: e(k) = 4e(k-1) + 5, for all integers k >= 1 e(0) = 2 **All terms in parenthesis are subscripts

## Best Answers: Use iteration to guess an explicit formula for the sequence?

Aubrey | 4 days ago
e(k) = 4e(k-1) + 5 = 4(4e(k-2) + 5) + 5 = 4^2 e(k-2) + 25 = 4^2 (4e(k-3) + 5) + 25 = 4^3 e(k-3) + 105 = ... It's obvious that the this will be of the form 4^j e(k-j) + c(j) for some constant c(j). We have the first few values of c(j): c(1) = 5 c(2) = 4*5 + 5 = 25 c(3) = 4*4*5 + 4*5 + 5 = 105 ... From this, it's pretty clear that c(j) = 4^(j-1)*5 + 4^(j-2)*5 + ... + 4^(j-j)*5 This can be evaluated more easily using the geometric series formula. Here, it gives c(j) = 5/3 * (4^j - 1) So, we have e(k) = 4^j e(k-j) + 5/3 * (4^j - 1) Setting j=k and using e(0) = 2 gives the final formula, e(k) = 2 * 4^k + 5/3 * (4^k - 1) = (11 * 4^k - 5) / 3 Of course this can be proven with an easy induction. P.S. Often, subscripts are typed like this: e_k, or e_k-1, or e_{k-1}. The first and third options are valid LaTeX code, which is the typesetting system that's used very heavily in math and related fields.
👍 136 | 👎 4
Did you like the answer? Use iteration to guess an explicit formula for the sequence? Share with your friends

We found more questions related to the topic: Types of parenthesis in math

Abaigael
Could you help me with a problem similar to this? I can figure out the first few answers but not the explicit formula.
👍 50 | 👎 -2