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

### Step 1: Identify the number of perfect squares up to a certain integer.
The perfect squares are numbers like \(1^2, 2^2, 3^2, \ldots, n^2\). The largest perfect square less than or equal to a number \(N\) is given by \(k^2\), where \(k\) is the largest integer such that \(k^2 \leq N\).

### Step 2: Determine how many perfect squares are there up to a certain number.
The number of perfect squares less than or equal to \(N\) is equal to \(k\), where \(k = \lfloor \sqrt{N} \rfloor\).

### Step 3: Relate the new sequence to the original sequence.
If we denote the original sequence of positive integers as \(1, 2, 3, \ldots, N\), then the new sequence (after removing perfect squares) will have \(N - k\) terms, where \(k\) is the number of perfect squares up to \(N\).

### Step 4: Set up the equation to find \(N\).
We want to find \(N\) such that:
\[ N - \lfloor \sqrt{N} \rfloor = 2015 \]
This means that the number of terms remaining after removing perfect squares equals 2015.

### Step 5: Estimate \(N\).
Let’s assume \(N\) is around 2060 (since we will add the number of perfect squares back later). We can calculate:
\[ \lfloor \sqrt{2060} \rfloor = 45 \quad \text{(since } 45^2 = 2025 \text{ and } 46^2 = 2116\text{)} \]
Now, substituting back:
\[ 2060 - 45 = 2015 \]
This confirms that when \(N = 2060\), we have exactly 2015 terms left in the new sequence.

### Step 6: Conclusion.
Thus, the 2015th term in the new sequence (after removing all perfect squares) is:
\[ \boxed{2060} \]