Animated Solution for Mathematics - Sets and Relations: Let $A=\{n\in \mathbb{N}\mid n^2\le n+10,000\}$, $B=\{3k+1\mid k\in \mathbb{N}\}$ and $C=\{2k\mid k\in \mathbb{N}\}$, then the sum of all the elements of the set $A\cap (B-C)$ is equal to

We start with the set $A=\{n\in \mathbb{N}\mid n^2\le n+10,000\}$. To simplify this, we rearrange the inequality to:

Factoring the expression, we obtain $n(n-1)\le 10,000$. We are looking for the product of two consecutive integers that does not exceed $10,000$.

Taking the square root of $10,000$ gives $100$. Testing the boundary, if $n=100$, then $100\times 99=9,900$, which is valid. If $n=101$, then $101\times 100=10,100$, which exceeds the threshold. Thus, Set $A$ consists of natural numbers from $1$ to $100$.

Next, we consider $B=\{3k+1\mid k\in \mathbb{N}\}$ and $C=\{2k\mid k\in \mathbb{N}\}$. The operation $B-C$ removes all even numbers from $B$.

For an element $3k+1$ to be odd, $3k$ must be even. Since $3$ is odd, $k$ must be even. Let $k=2m$, where $m\in \mathbb{N}$.

Substituting this into the general term, we get:

This confirms that the elements of $B-C$ are of the form $6m+1$.

We now find the intersection $A\cap (B-C)$. Given $A=\{1,2,\dots ,100\}$ and $B-C=\{7,13,19,\dots \}$, we require:

Since $m$ must be a natural number, $m\in \{1,2,\dots ,16\}$. This forms an Arithmetic Progression (AP) where the first term $a=6(1)+1=7$ and the last term $l=6(16)+1=97$.

The number of terms in this progression is $n=16$.

The sum of an AP is given by the formula:

Substituting our identified values into the equation:

The final sum is 832.