The weighted AM of first n natural numbers whose weights are equal to the corresponding numbers is equal to
2n+1
12(2n+1)
13(2n+1)
2n+16
The required weighted mean is given by
X¯=1⋅1+2⋅2+3⋅3+…+n⋅n1+2+3+…+nX¯=n(n+1)(2n+1)6n(n+1)2=2n+13