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house of the dragon season 1 - Network4

📅 May 19, 2026 👤 scraface

May 19, 2026

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🔥 Popular Posts

But sequence 4–8: 4 (div by 4), 6 (div by 2), 8 (div by 8) → divisible...
Even sequence 5–9: 8 → div by 8.
Actually, in any five consecutive integers, there is always a multiple...
Similarly, one divisible by 3, and one divisible by 5.
Thus, the product is divisible by \( 8 \times 3 \times 5 = 120 \).
Is there a larger fixed divisor?
Check divisibility by 7: not guaranteed (e.g., 1–5: no 7) → so 7 does...
Check powers: is it divisible by \( 16 \)? Try 1–5: product \( 120 \),...
Check 9: need two multiples of 3, but in five consecutive, only one mu...
Any five consecutive integers must include at least one multiple of 3?...

📝 Recent Posts

\lim_{n \to \infty} a_n = \lim_{n \to \infty} \sum_{k=1}^{n} \left( \f...
This is a Riemann sum. Split the sum:
a_n = \sum_{k=1}^{n} \frac{k}{n} - \frac{1}{3} \sum_{k=1}^{n} \left( \...
= \frac{1}{n} \sum_{k=1}^{n} k - \frac{1}{3n^3} \sum_{k=1}^{n} k^3
Use formulas:
$ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} $
$ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 $
a_n = \frac{1}{n} \cdot \frac{n(n+1)}{2} - \frac{1}{3n^3} \cdot \left(...
= \frac{n+1}{2} - \frac{1}{3n^3} \cdot \frac{n^2(n+1)^2}{4}
Simplify the second term:

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