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Home » \frac1n(n+2) = \fracAn + \fracBn+2. - jntua results

\frac1n(n+2) = \fracAn + \fracBn+2. - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 Subtract \( k \) from both sides:

📌 = 8k \quad \Rightarrow \quad k = 0

📌 A science fiction writer models the growth of a resource-dependent colony on Mars using the equation \( a + b = 12 \) and \( a^2 + b^2 = 80 \). Find \( a^3 + b^3 \).

📌 a + b = 12 \quad \text{and} \quad a^2 + b^2 = 80

📌 Use the identity:

📌 We know \( (a + b)^3 = 12^3 = 1728 \). To use the identity, we need \( ab \). From the identity:

📌 = 12^2 - 2ab = 144 - 2ab

📌 Solve for \( ab \):

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Subtract: \( 2,400,000 < 120,000t \)
Divide: \( t > 2,400,000 / 120,000 = 20 \)
So after 21 years, it drops below. But the question asks when it *drop...
But numerically: \( t = 20 \) gives exactly 6M → needs to go below, so...
However, math solution: \( t > 20 \), so after 20 years it reaches 6M,...
Solve: \( 8,400,000 - 120,000t < 6,000,000 \Rightarrow t > 20 \) → so...
But precise answer: the number of years is 21.
However, standard is to report the integer value satisfying, so t = 21...
But let's compute:
At t=20: 8.4M - 1.2M×20 = 8.4M - 24M = negative? No: 120,000 × 20 = 2,...

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