The equal resistance of 'n' resistors each of Resistance 'R' , Rs and Rp in series and parallel , prove that (Rs - Rp) = (n square - 1)R/n

Solution: 

We know that equivalent resistance in series connection is sum of all the individual resistances.

So, In series connection R_s = nR

Also The equivalent resistance in parallel connection is

1/R_p = 1/R+1/R+1/R+...................n times

=> 1/R_p = (1+1+1+1+........n times)/R

=> R_p = R/n

In parallel connection R_p = R/n

Now R_s - R_p = nR-R/n

=> R_s - R_p = (n²R-R)/n

=> R_s - R_p = (n^​2​- 1)R/n  Proved

Solution: 

We know that equivalent resistance in series connection is sum of all the individual resistances.

So, In series connection R_s = nR  and 

In parallel connection R_p = R/n

Now R_s - R_p = nR-R/n

=> R_s - R_p = (n²R-R)/n

=> R_s - R_p = (n^​2​- 1)R/n  Proved