akhaliq/anychat · DeepSeek V3 Sovling recurrence

😃: if S(2^(n+1)) = 2*S(2^n) + 2^n for n=1,2,3,4 ...
and S(2) =1
find a general value for S(2^n)
🤖: We are given the recurrence relation:

[
S(2^{n+1}) = 2 \cdot S(2^n) + 2^n
]

with the initial condition ( S(2) = 1 ). We want to find a general expression for ( S(2^n) ).

Step 1: Rewrite the recurrence

Let ( a_n = S(2^n) ). Then the recurrence becomes:

[
a_{n+1} = 2a_n + 2^n
]

with the initial condition ( a_1 = 1 ).

Step 2: Solve the recurrence

This is a linear nonhomogeneous recurrence relation. We first solve the homogeneous part:

[
a_{n+1}^{(h)} = 2a_n^{(h)}
]

The solution to this is:

[
a_n^{(h)} = C \cdot 2^n
]

Next, we find a particular solution to the nonhomogeneous recurrence. We guess a solution of the form ( a_n^{(p)} = A \cdot n \cdot 2^n ). Substituting into the recurrence:

[
A \cdot (n+1) \cdot 2^{n+1} = 2 \cdot A \cdot n \cdot 2^n + 2^n
]

Simplify:

[
2A(n+1)2^n = 2An2^n + 2^n
]

Divide both sides by ( 2^n ):

[
2A(n+1) = 2An + 1
]

Simplify:

[
2An + 2A = 2An + 1
]

This gives:

[
2A = 1 \implies A = \frac{1}{2}
]

Thus, the particular solution is:

[
a_n^{(p)} = \frac{1}{2} \cdot n \cdot 2^n = n \cdot 2^{n-1}
]

The general solution is the sum of the homogeneous and particular solutions:

[
a_n = C \cdot 2^n + n \cdot 2^{n-1}
]

Step 3: Apply the initial condition

Using ( a_1 = 1 ):

[
1 = C \cdot 2^1 + 1 \cdot 2^{0} = 2C + 1
]

Solving for ( C ):

[
2C = 0 \implies C = 0
]

Thus, the solution is:

[
a_n = n \cdot 2^{n-1}
]

Step 4: Write the final expression

Since ( a_n = S(2^n) ), we have:

[
S(2^n) = n \cdot 2^{n-1}
]

Verification

For ( n = 1 ):

[
S(2^1) = 1 \cdot 2^{0} = 1
]

For ( n = 2 ):

[
S(2^2) = 2 \cdot 2^{1} = 4
]

For ( n = 3 ):

[
S(2^3) = 3 \cdot 2^{2} = 12
]

These values satisfy the recurrence relation, confirming the solution.

Final Answer

[
S(2^n) = n \cdot 2^{n-1}
]