the non-

decreasing sequence of odd integers {a1, a2, a3, .
. .} = {1,3,3,3,5,5,5,5,5,...} each positive odd
integer k appears k times. it is a fact that there
are integers b, c, and d such that, for all positive
integers n, añ = b[√(n+c)] +d. Where [x] denotes the largest integer not exceeding x. Find the sum of b+c+d.

Show step too hard

To find the sum of b+c+d, we need to determine the values of b, c, and d first. Let's start by analyzing the given non-decreasing sequence of odd integers {a1, a2, a3, ...} = {1, 3, 3, 3, 5, 5, 5, 5, 5, ...}.

First Pattern:
Looking closely at the sequence, we can observe that each positive odd integer k appears k times. For example, 1 appears once, 3 appears three times, 5 appears five times, and so on. This pattern helps us generate the sequence.

Second Pattern:
Next, let's analyze the formula given for the nth term - añ = b[√(n+c)] + d. Since the sequence we are trying to generate is non-decreasing, we know that añ should always be greater than or equal to the previous term. Let's consider the values of n for each position in the sequence.

For n = 1, añ (first term) = b√(1+c) + d
For n = 2, añ (second term) = b√(2+c) + d
For n = 3, añ (third term) = b√(3+c) + d

By comparing these terms, we notice that when the value of n increases, the value of √(n+c) also increases. This suggests that as n increases, the value inside the square root might also increase linearly.

Now, let's consider the first pattern observed in the sequence: each positive odd integer k appears k times. This means that each position of the sequence represents a specific odd number. We can express this relationship as:

For n = 1, añ (first term) = 1
For n = 2, añ (second term) = 3
For n = 3, añ (third term) = 3
For n = 4, añ (fourth term) = 3
For n = 5, añ (fifth term) = 5
For n = 6, añ (sixth term) = 5

From this, we can deduce that the positions of the sequence, when arranged in ascending order, represent the odd numbers in a non-decreasing manner. This suggests that the value inside the square root is related to the position of the term.

Given these two patterns, we can conclude that n+c is equal to the position of the term in the sequence. Therefore, n+c = k, where k represents the position of the term.

Now, let's summarize the relationships we have discovered so far:

For n = k, añ (kth term) = b√k + d
n + c = k

By substituting n + c = k in the formula, we get:

añ = b√(n+c) + d
añ = b√k + d

Therefore, b√k + d represents the kth term in the sequence.

From the first pattern, we know that each positive odd integer k appears k times in the sequence. So, the sum of the first k terms can be calculated as follows:

Sum of the first k terms = k * (b√k + d)

To find the sum of the entire sequence, we need to calculate the sum of the terms up to infinity. We need to find the value of b+c+d, which will give us the sum of the terms.

Unfortunately, with the given information, we cannot determine the exact values of b, c, and d, as there are infinite possibilities that could satisfy the given conditions. Without additional constraints or information, it is impossible to determine the sum of b+c+d.

since the terms of the sequence are all odd numbers (2k+1), and there are k^2 terms <= the kth odd number, look for something like

2[√n]+1

That sequence is

3,3,3, 5,5,5,5,5,5,5, ...

Looks like we have to adjust it by 1 to get

2[√(n-1)]+1

So, b+c+d = 2

No, I don't have an actual proof, but maybe you can work from here. I'll think on it some.