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A sequence $u_1, u_2, u_3, \ldots$ is defined as follows, for $n \in \mathbb{N}$: $u_1 = 2, \quad u_2 = 64, \quad u_{n+1} = \frac{u_n}{\sqrt{u_{n-1}}}$ Write $u_3$ in the form $2^p$, where $p \in \mathbb{R}$ - Leaving Cert Mathematics - Question 4 - 2022

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A-sequence-$u_1,-u_2,-u_3,-\ldots$-is-defined-as-follows,-for-$n-\in-\mathbb{N}$:--$u_1-=-2,-\quad-u_2-=-64,-\quad-u_{n+1}-=-\frac{u_n}{\sqrt{u_{n-1}}}$--Write-$u_3$-in-the-form-$2^p$,-where-$p-\in-\mathbb{R}$-Leaving Cert Mathematics-Question 4-2022.png

A sequence $u_1, u_2, u_3, \ldots$ is defined as follows, for $n \in \mathbb{N}$: $u_1 = 2, \quad u_2 = 64, \quad u_{n+1} = \frac{u_n}{\sqrt{u_{n-1}}}$ Write $u_3$... show full transcript

Worked Solution & Example Answer:A sequence $u_1, u_2, u_3, \ldots$ is defined as follows, for $n \in \mathbb{N}$: $u_1 = 2, \quad u_2 = 64, \quad u_{n+1} = \frac{u_n}{\sqrt{u_{n-1}}}$ Write $u_3$ in the form $2^p$, where $p \in \mathbb{R}$ - Leaving Cert Mathematics - Question 4 - 2022

Step 1

Write $u_3$ in the form $2^p$

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Answer

Starting with the recursive definition:

  1. First, we need to calculate u3u_3 using the provided definitions:

    • We already have that:

    u1=2u_1 = 2 u2=64=26u_2 = 64 = 2^6

    • Now, compute u3u_3:

    u3=u2u1=642u_3 = \frac{u_2}{\sqrt{u_1}} = \frac{64}{\sqrt{2}}

    • Simplifying this expression:

    u3=2621/2=261/2=25.5u_3 = \frac{2^6}{2^{1/2}} = 2^{6 - 1/2} = 2^{5.5}

    • Therefore, we can write u3u_3 as:

    u3=2112u_3 = 2^{\frac{11}{2}}

Step 2

By letting $y = e^k$ in this arithmetic sequence, show that: $5y^2 - 26y + 5 = 0$

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Answer

In this arithmetic sequence, we start with:

  • The first term is 5ek5e^{-k}, the second term is 1313, and the third term is 5ek5e^{k}.

  • Recognizing that these terms must maintain a common difference, we set:

T2T1=T3T2T_2 - T_1 = T_3 - T_2

  • Let’s substitute the terms:

135ek=5ek1313 - 5e^{-k} = 5e^{k} - 13

  • Rearranging gives:

5ek+5ek=265e^{-k} + 5e^{k} = 26

  • Multiplying through by eke^k (to eliminate eke^{-k}):

5+5e2k=26ek5 + 5e^{2k} = 26e^{k}

  • Rearranging the equation:

5e2k26ek+5=05e^{2k} - 26e^{k} + 5 = 0

  • Substituting y=eky = e^k leads to:

5y226y+5=05y^2 - 26y + 5 = 0

Step 3

Use the equation in $y$ in part (b)(i) to find the two possible values of $k$

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Answer

To solve the quadratic equation 5y226y+5=05y^2 - 26y + 5 = 0, we can use the quadratic formula:

y=b±b24ac2ay = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where a=5a = 5, b=26b = -26, and c=5c = 5. Plugging in the values:

  • The discriminant is:

b24ac=(26)24(5)(5)=676100=576b^2 - 4ac = (-26)^2 - 4(5)(5) = 676 - 100 = 576

  • Therefore, we have:

y=26±57625y = \frac{26 \pm \sqrt{576}}{2 \cdot 5}

  • Simplifying further:

y=26±2410y = \frac{26 \pm 24}{10}

  • This gives two solutions:
  1. y1=5010=5y_1 = \frac{50}{10} = 5
  2. y2=210=15y_2 = \frac{2}{10} = \frac{1}{5}
  • Recalling y=eky = e^k, we get:
  1. For y1=5y_1 = 5:
    • ek=5k=ln(5)e^k = 5 \Rightarrow k = \ln(5)
  2. For y2=15y_2 = \frac{1}{5}:
    • ek=15k=ln(15)=ln(5)e^k = \frac{1}{5} \Rightarrow k = \ln\left(\frac{1}{5}\right) = -\ln(5)
  • Thus, the two possible values of kk in the required form are:

  • k=ln(5)k = \ln(5) and k=ln(5)k = -\ln(5).

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