A new sequence is obtained from the sequence of positive integers `1,2,……..,` by deleting all the perfect squares. What is the 2015th term from the beginning of the new sequence ?

Answer

To find the 2015th term of the new sequence obtained by deleting all perfect squares from the sequence of positive integers, we can follow these steps:

### Step 1: Understand the sequence
The original sequence is the set of positive integers: \(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \ldots\)

The perfect squares in this sequence are: \(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, 3969, 4096, 4225, 4356, 4489, 4624, 4761, 4900, 5041, 5184, 5329, 5476, 5625, 5764, 5929, 6084, 6241, 6400, 6561, 6724, 6889, 7056, 7225, 7396, 7569, 7744, 7921, 8100, 8281, 8464, 8649, 8836, 9025, 9216, 9409, 9604, 9801, 10000\)

### Step 2: Count the perfect squares up to a certain number
We need to determine how many perfect squares exist up to a certain integer \(n\). The number of perfect squares less than or equal to \(n\) is given by \(\lfloor \sqrt{n} \rfloor\).

### Step 3: Set up the equation
If we denote \(n\) as the term we want to find in the original sequence, then the number of terms remaining in the new sequence after removing the perfect squares is:
\[ \text{Remaining terms} = n - \lfloor \sqrt{n} \rfloor \]

We want this to equal 2015:
\[ n - \lfloor \sqrt{n} \rfloor = 2015 \]

### Step 4: Solve for \(n\)
Rearranging gives:
\[ n = 2015 + \lfloor \sqrt{n} \rfloor \]

### Step 5: Estimate \(n\)
Let \(k = \lfloor \sqrt{n} \rfloor\). Then we can express \(n\) as:
\[ n = 2015 + k \]
Since \(k\) is the integer part of the square root of \(n\), we can substitute:
\[ k = \lfloor \sqrt{2015 + k} \rfloor \]

### Step 6: Approximate \(k\)
Assuming \(k\) is around 44 (since \(44^2 = 1936\) and \(45^2 = 2025\)), we can check:
\[ n = 2015 + 44 = 2059 \]

### Step 7: Check the perfect squares
Now we calculate the number of perfect squares up to 2059:
\[ \lfloor \sqrt{2059} \rfloor = 45 \]
Thus:
\[ \text{Remaining terms} = 2059 - 45 = 2014 \]

Since we need 2015 terms, we try \(n = 2060\):
\[ \lfloor \sqrt{2060} \rfloor = 45 \]
\[ \text{Remaining terms} = 2060 - 45 = 2015 \]

### Conclusion
Thus, the 2015th term in the new sequence is the 2060th term in the original sequence of positive integers.

**Final Answer: 2060**