#C++, 53 bytes

Just use the formula. n is 1-indexed and k is 0-indexed.

[](int n,int k){return k/n*(k/n+1)/2*n+k%n*(k/n+1);};

Try it online!

C++, 53 bytes

Just use the formula. n is 1-indexed and k is 0-indexed.

[](int n,int k){return k/n*(k/n+1)/2*n+k%n*(k/n+1);};

Try it online!

#C++, 53 bytes

Just use the formula. n is 1-indexed and k is 0-indexed.

[](int n,int k){return k/n*(k/n+1)/2*n+k%n*(k/n+1);};

Try it online!

Use exactly the same approach on C will get 43 bytes, and on Python 2 38 bytes.

#C++, 53 bytes

Just use the formula. n is 1-indexed and k is 0-indexed.

[](int n,int k){return k/n*(k/n+1)/2*n+k%n*(k/n+1);};

Try it online!

#C++, 53 bytes

Just use the formula. n is 1-indexed and k is 0-indexed.

[](int n,int k){return k/n*(k/n+1)/2*n+k%n*(k/n+1);};

Try it online!

#C++, 53 bytes

Just use the formula. n is 1-indexed and k is 0-indexed.

[](int n,int k){return k/n*(k/n+1)/2*n+k%n*(k/n+1);};

Try it online!

Use exactly the same approach on C will get 43 bytes, and on Python 2 38 bytes.