10 quesitons together

Prove that for all integers $n \geq 1$, the expression $5^n – 1$ is divisible by 4.

That is, show that $4 \mid (5^n – 1)$ for all $n \in \mathbb{N}$.

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Step 1: Base Case (n = 1)

When $n = 1$, we have $5^1 – 1 = 4$, which is divisible by 4.
✅ The base case holds.

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Step 2: Inductive Hypothesis

Assume that for some $k \geq 1$, the expression $5^k – 1$ is divisible by 4.
That is, assume $5^k – 1 = 4m$ for some integer $m$.
This implies that $5^k \equiv 1 \pmod{4}$.

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Step 3: Inductive Step

We want to prove that $5^{k+1} – 1$ is also divisible by 4.

We can write $5^{k+1} – 1 = 5 \cdot 5^k – 1$.

Since $5^k \equiv 1 \pmod{4}$ (by the inductive hypothesis), we get
$5 \cdot 5^k \equiv 5 \cdot 1 = 5 \equiv 1 \pmod{4}$.

Therefore, $5^{k+1} – 1 \equiv 0 \pmod{4}$, which means $4 \mid (5^{k+1} – 1)$.

✅ The inductive step is proven.

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Conclusion:

By the principle of mathematical induction, we conclude that $5^n – 1$ is divisible by 4 for all integers $n \geq 1$.

$ \boxed{\text{Q.E.D.}} $

<p>
<strong>Step 1 (Base Case):</strong> For \(n = 1\), the left-hand side is \(1\) and the right-hand side is \(1^2 = 1\), so the statement holds.<br>
<strong>Step 2 (Inductive Hypothesis):</strong> Assume for some \(k \ge 1\) that
\(1 + 3 + 5 + \dots + (2k – 1) = k^2\).<br>
<strong>Step 3 (Inductive Step):</strong> Then for \(n = k+1\),
\[
1 + 3 + \dots + (2k-1) + \bigl(2(k+1)-1\bigr)
= k^2 + (2k + 1)
= (k+1)^2.
\]
So the statement holds for \(k+1\).<br>
<strong>Final:</strong> By induction, for all \(n \ge 1\),
\(\boxed{1 + 3 + 5 + \dots + (2n – 1) = n^2}\)
</p>

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The first three terms of an arithmetic sequence are

$u_1$,

$3u_1 – 9$,

$2u_1 + 3$.
(a) Show that

$u_1 = 7$.
(b) Find the common difference

$d$.
(c) Which term of the sequence is 102?
(d) Find the sum of the first 20 terms of the sequence.

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(a) Since the sequence is arithmetic, compute the two consecutive differences:
Compute

$(3u_1 – 9) – u_1 = 2u_1 – 9$and

$(2u_1 + 3) – (3u_1 – 9) = -u_1 + 12$,
equate

$2u_1 – 9 = -u_1 + 12$to get

$3u_1 = 21$and hence

$u_1 = 7$.

$\boxed{u_1 = 7}$ 

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(b) Compute the common difference:
Compute

$d = (3u_1 – 9) – u_1 = 2u_1 – 9 = 2\cdot7 – 9 = 5$.

$\boxed{d = 5}$ 

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(c) Find which term is 102:
Use

$u_n = u_1 + (n-1)d = 7 + 5(n-1)$, set

$7 + 5(n-1) = 102$giving

$5n + 2 = 102$so

$n = 20$.

$\boxed{n = 20}$ 

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(d) Sum of the first 20 terms:
Use

$S_{20} = \tfrac{20}{2}(u_1 + u_{20}) = 10\,(7 + 102) = 1090$.

$\boxed{S_{20} = 1090}$ 

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