Q.2
. Find the product of
101*10001*100000001*...*(1000...01) where the last factor has _{} zeroes between the ones. Find the number of
ones in the product.

**Solution **

We know,

101 x 10001 = 1010101

10001 has 2^2
-1 zeroes between the ones.

The product has 2^2-1 zeroes
alternately placed between 2^2 ones. It starts with one and ends with one.

1010101 x 100000001 = 101010101010101

100000001 has
2^3 -1 zeroes between the ones.

The product has 2^3-1 zeroes
alternately placed between 2^3 ones. It starts with one and ends with one.

This multiplication pattern shows
that, for such a multiplication series till 1000...001 with 2^n-1 zeroes, the
product will have 2^n-1 zeroes alternately

placed
between 2^n ones.(As above)

Therefore the product of
the series till 1000...001 having 2^7-1 zeroes would be,

10101010........01 with 2^7-1 zeroes alternately placed between
2^7 ones.