Solution:
We know that equivalent resistance in series connection is sum of all the individual resistances.
So, In series connection Rs = nR
Also The equivalent resistance in parallel connection is
1/Rp = 1/R+1/R+1/R+...................n times
=> 1/Rp = (1+1+1+1+........n times)/R
=> Rp = R/n
In parallel connection Rp = R/n
Now Rs - Rp = nR-R/n
=> Rs - Rp = (n2R-R)/n
=> Rs - Rp = (n2- 1)R/n Proved