I think answer is (4)

p is surely $2^{k+1} - 1$ .. bcoz terms are added in this order - 1,2,4,8,16.. at any time there sum is next power of 2 minus 1.

k should be less than m. If k=m, then corresponding p value will not hold.

p is surely $2^{k+1} - 1$ .. bcoz terms are added in this order - 1,2,4,8,16.. at any time there sum is next power of 2 minus 1.

k should be less than m. If k=m, then corresponding p value will not hold.