Question

What is the ratio of equivalent resistance of series combination of n equal resistance to equivalent resistance in parallel combination of those n resistance?

Answer 1

The answer to your question is quite simple. Let me explain it from the scratch. 

If two resistances are connected in series like in the figure.
Lets assume, the voltage applied is V and current flowing in the circuit is I. In series combination, the current is same but the voltage applied will be divided among R1 and R2. Lets assume the voltage drop across R1 is V1 and R2 is V2. Now V=V1+V2. We also know from ohms law that V=IR. Therefore, V1=IR1 and V2=IR2. NOW 
V=V1+V2
Since current flowing through the circuit is same we can write
IRseq=IR1+IR2
In your case, both resistances are equal so R1=R2=R
Rseq = R+R = 2R


Now connecting them in parallel..
The voltage across the resistances is same as the applied voltage V but the current divides. Lets say the current through R1 is I1 and through R2 is I2 and the totally current drawn from battery is I
So I = I1+I2
And from ohms law I =V/R
So V/Rpeq = V/R1 + V/R2
which leaves 1/Rpeq = 1/R1 + 1/R2
Again the two resistances are equal R1=R2=R
So Rpeq = R/2

Now the ration of Series equivalent resistance to Parallel equivalent resistance will be 
Rseq/Rpeq = 2R/(R/2) = 4

If it is N number of resistances, then the ration will N^2 as everybody else said. In your case N=2, i.e. 4 is the answer.
I guess it helps you...

Answer 2

R = 2 R. ...
As both are in series so current will be same while voltage canews and R1 = R2 so R = 2 R
But in this question there is no resistance so R = nR