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
Question 4
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:
First, we need to calculate u3 using the provided definitions:
We already have that:
u1=2u2=64=26
Now, compute u3:
u3=u1u2=264
Simplifying this expression:
u3=21/226=26−1/2=25.5
Therefore, we can write u3 as:
u3=2211
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 5e−k, the second term is 13, and the third term is 5ek.
Recognizing that these terms must maintain a common difference, we set:
T2−T1=T3−T2
Let’s substitute the terms:
13−5e−k=5ek−13
Rearranging gives:
5e−k+5ek=26
Multiplying through by ek (to eliminate e−k):
5+5e2k=26ek
Rearranging the equation:
5e2k−26ek+5=0
Substituting y=ek leads to:
5y2−26y+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 5y2−26y+5=0, we can use the quadratic formula:
y=2a−b±b2−4ac
where a=5, b=−26, and c=5. Plugging in the values:
The discriminant is:
b2−4ac=(−26)2−4(5)(5)=676−100=576
Therefore, we have:
y=2⋅526±576
Simplifying further:
y=1026±24
This gives two solutions:
y1=1050=5
y2=102=51
Recalling y=ek, we get:
For y1=5:
ek=5⇒k=ln(5)
For y2=51:
ek=51⇒k=ln(51)=−ln(5)
Thus, the two possible values of k in the required form are:
k=ln(5) and k=−ln(5).
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