x_n = number of binary strings of length n that contains non-consecutive 0’s.
⇒ Let us consider n=1
Possible binary strings: 0, 1
So, x₁ = 2
⇒ For n = 2;
Possible strings are,
010
110
111
011
101
So, x₃ = 5
⇒ For n=4,
Possible strings are
0 1 1 1
1 0 1 1
1 1 0 1
1 1 1 0
0 1 1 0
0 1 0 1
1 0 1 0
1 1 1 1

So,
x₄=8,
Hence, for all values of x₄ above only option (D) satisfies.

From above question we have found that equation,
x_n = x_n-1 + x_n-2
satisfies.
And also we have found the value of x₄ and x₃ for the above equation.
So, according to the equation the value of x₅ will be,
x₅ = x₃ + x₄
= 5 + 8
= 13
Hence , none of the given option is correct.

T(1) =1
T(n) = 2T(n-1) + n
T(2) = 2T(1) + 2 = 2 + 2 = 4
T(3) = 2T(2) + n = 2(4) + 3 = 11
T(4) = 2T(3) + 4 = 22 + 4 = 26
Let check with the options:
Option A:
n=4
2⁴⁺¹ - 4 - 2
32 - 6
26 (✔️)
Option B:
n=4
2ⁿ-n
2⁴-4
12(✖️)
Option C:
n=4
2⁴⁺¹ - 2(4) - 8
32 - 10
22(✖️)
Option D:
n=4
2ⁿ - n
2⁴ - 4
12(✖️)

Considering floor value for square root of numbers.
Successive root number of numbers are in the series 3, 5, 7, ... like 3 numbers from 1... 4, 5 numbers 5-9 and so on.

T(2^k)=3T(2^k-1)+1
T(1)=1
k=0; T(1)=3T(2^-1)+1
k=1; T(2)=3T(2⁰)+1=3(1)+1=4
k=2; T(4)=3T(2¹)+1=3(4)+1=13
k=3; T(8)=3T(2²)+1=3(13)+1=40
k=4; T(16)=3T(2³)+1=3(40)+1=121
Simply go through the options.
Option B:
k=4 ⇒ (3⁴⁺¹-1)/2
⇒ (243-1)/2
⇒ 121