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Find a general formula for Un for the sequence defined by Un+1 = 3 + rac14 Un with U1 = 5.

Mathematics

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Answer

Un=14n−1+4U_n = \frac{1}{4^{n-1}} + 4

Step-by-step solution

5 steps

It looks like you're asking to find a general formula for UnU_n for the sequence defined in the original image: Un+1=3+14UnU_{n+1} = 3 + \frac{1}{4} U_n with U1=5U_1 = 5. This is often the next step after calculating the first few terms.

Here's how to find the formula for UnU_n using the general method:

  1. Find the homogeneous solution.

    The homogeneous part of the recurrence relation is Un+1=14UnU_{n+1} = \frac{1}{4} U_n.
    The characteristic equation is r=14r = \frac{1}{4}.
    The homogeneous solution is Un(h)=A(14)nU_n^{(h)} = A \left(\frac{1}{4}\right)^n, where AA is a constant.

  2. Find the particular solution.

    Assume a constant particular solution Un(p)=CU_n^{(p)} = C. Substitute this into the original recurrence relation:

    C=3+14CC = 3 + \frac{1}{4} C

    Subtract 14C\frac{1}{4} C from both sides:

    C−14C=3C - \frac{1}{4} C = 3 34C=3\frac{3}{4} C = 3

    Multiply by 43\frac{4}{3}:

    C=3×43C = 3 \times \frac{4}{3} C=4C = 4

    So, the particular solution is Un(p)=4U_n^{(p)} = 4.

  3. Combine to find the general solution.

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

    Un=Un(h)+Un(p)U_n = U_n^{(h)} + U_n^{(p)} Un=A(14)n+4U_n = A \left(\frac{1}{4}\right)^n + 4
  4. Determine the constant AA using the initial term U1=5U_1 = 5.

    Substitute n=1n=1 and U1=5U_1 = 5 into the general solution:

    5=A(14)1+45 = A \left(\frac{1}{4}\right)^1 + 4 5=A4+45 = \frac{A}{4} + 4

    Subtract 4 from both sides:

    1=A41 = \frac{A}{4}

    Multiply by 4:

    A=4A = 4
  5. Substitute AA back into the general solution.

    Un=4(14)n+4U_n = 4 \left(\frac{1}{4}\right)^n + 4

    This can be simplified:

    Un=44n+4U_n = \frac{4}{4^n} + 4 Un=14n−1+4U_n = \frac{1}{4^{n-1}} + 4

    The formula for UnU_n is Un=14n−1+4\boxed{U_n = \frac{1}{4^{n-1}} + 4}.

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