5710 Shares

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

July 19, 2019 / By Chonsie

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

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

The following polynomial p(x) works: (I have edited these since first posting due to a small error. The actual values don't really matter though, only that the technique works.) p(x) = R(x) + i * i(x) where R(x) ≈ -0.0043663095238095*x^9 + 0.20872845238095*x^8 - 4.261379523809527*x^7 + 48.57609277777789*x^6 - 339.1883991666676*x^5 + 1496.427338611115*x^4 - 4142.960431190484*x^3 + 6878.457840158739*x^2 - 6135.313423809527*x + 2199.058 i(x) ≈ .00044137557870370375*x^9 - 0.020268647073413*x^8 + 0.39863961061508*x^7 - 4.393082552083342*x^6 + 29.77215699131946*x^5 - 128.0045194010416*x^4 + 346.7821787962959*x^3 - 565.647609399801*x^2 + 497.6784332261902*x - 176.56637 These are approximate as π was approximated as well as certain coefficients. [ i(x) has zeros at x = 1,2,3,...,8 ] Simply create two polynomials, one real, one imaginary, and using whatever technique you wish subject to: for x=1,2,3,...,10; R(x) has values: 1, 2, 3, 4, 5, 7, 24.6, -0.1124, 1, -2 for x=1,2,3,...,10; i(x) has values: 0, 0, 0, 0, 0, 0, 0, 0, -16.4, 4π No discontinuities anywhere.

Wow! What a concept! I never thought of this for my self but I would have to say YES. I would go for it. I am the single parent of a thirteen year old, having being on this side of the fence for so long I would take any help I can get. Every child needs the benefits of two loving parents. Currently I keep a clean home, do laundry, pay bills, So if I had someone to take all of that pressure off of me then why not? I make a considerable amount of money enough to comfortably support a family of three so if I had just a little support... It's all good.

If you have your own answer to the question types of parenthesis in math, then you can write your own version, using the form below for an extended answer.