# Solve the recursive relation by finding a closed-form formula and vice versa?

Q1. Find the closed-form formula for c_k:

c_k=c_(k-1) + 2k, where c_0 = 3 for k = 1, 2, 3, 4, ...

Q2. A_n = (A_n-1)[n/(n+2)]^(-1/2), A_1 = P/sqrt(2), find the closed-form formula.

Q3. Find a recursive relation with the given closed-formed formula

A = P/sqrt[(n(n+1)] for n = 2, 3, 4, 5, ...

Please show all the work, and please don't use guess and check if possible... what I want is a solid method in solving this.

In general, should there be exactly one closed-form formula for any recursive relation (is vice-versa also true)? What are some approaches in solvng this?

### 3 Answers

- Low Key LyesmithLv 51 decade agoFavorite Answer
Note that c_k - c_{k-1} = 2k. The sum of (c_k - c_{k-1}) from k=1 to n is (c_n - c_0). The sum of 2k from k=1 to n is n(n+1). Therefore, c_n - c_0 = n(n+1), so c_n = n^2 + n + 3.

The idea for Q2 is similar. The product of (A_k)/(A_{k-1}) from k=2 to n is A_n/A_1, and the product of [k/(k+1)]^(-1/2) from k=2 to n is [2/(n+1)]^(-1/2).

Both Q1 and Q2 use the idea of a telescoping series (although technically, Q2 is a telescoping product). I won't try to explain the method here, but http://en.wikipedia.org/wiki/Telescoping_series has some examples.

The solutions to Q1 and Q2 are unique. If we know c_0 then we can find c_1, and if we know c_1 then we can find c_2, and so on, so each c_n is determined uniquely. However, there could be different formulas that yield the same results.

Q3 has many possible solutions. One way to find a recurrence relation is to divide A_n by A_{n-1} and simplify.

- UnknownDLv 61 decade ago
Find c_k in terms of c_0.

c_k = c_0 + 2k + 2(k - 1) + 2(k - 2) + 2(k - 3) + ... 2

c_k = 3 + 2(1 + 2 + 3 + 4 + 5 + ... + k)

c_k = 3 + 2(k(k + 1) / 2)

c_k = k² + k + 3

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Here's the approach for Q1 and you will see that it works for Q2 and Q3. We currently have here a term expressed as the term before it. If we went far back enough, we will reach the term c_0, which has a known value of 3. Therefore we have lost an unknown. So the main point is to express everything into something you do know.

- luccaLv 43 years ago
locate c_k in terms of c_0. c_k = c_0 + 2k + 2(ok - a million) + 2(ok - 2) + 2(ok - 3) + ... 2 c_k = 3 + 2(a million + 2 + 3 + 4 + 5 + ... + ok) c_k = 3 + 2(ok(ok + a million) / 2) c_k = ok² + ok + 3 ----- right this is the attitude for Q1 and you will see that it extremely works for Q2 and Q3. We at present have right here a term expressed via fact the term till now it. If we went far lower back adequate, we can attain the term c_0, which has a common fee of three. hence we've lost an unknown. So the main efficient element is to particular each and everything into something you do comprehend.