Answer
$T_n = 2\times T_{n-1}$ ,
$T_1 = 1$
Work Step by Step
Let $T_n$ be the number of bit sequences of length $n$ with an even number of 0s.
∴ Number of bit sequences of length $n-1$ with an even number of 0s = $T_{n-1}$
Now a string with an even number of zeroes can start with either 0 or 1.
Number of strings that start with 1 = $T_{n-1}$
(no. of strings with an even number of 0s of length $n-1$)
Number of strings that start with 0 = $2^{n-1}-T_{n-1}$
(total number of strings of length $n-1$ minus those with even number of 0s)
So, $T_n = T_{n-1}+(2^{n-1}-T_{n-1}) = 2^{n-1}$
And $T_{n-1} = 2^{(n-1)-1} = 2^{n-2}$
Therefore , $T_n = 2\times T_{n-1}$
Initial condition: $T_1 = 1$, as the only bit strings of size 1 are 0 & 1 of which the latter has even(0) number of zeroes.
