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Home » = 3^3,\quad 64 = 4^3,\quad 125 = 5^3,\quad 216 = 6^3. - Veritas Home Health

= 3^3,\quad 64 = 4^3,\quad 125 = 5^3,\quad 216 = 6^3. - Veritas Home Health

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Solve for $ A, B, C $:
For $ k = 0 $: $ 1 = A(1)(2) \implies A = \frac{1}{2} $.
For $ k = -1 $: $ 1 = B(-1)(1) \implies B = -1 $.
For $ k = -2 $: $ 1 = C(-2)(-1) \implies C = \frac{1}{2} $.
\frac{1}{k(k+1)(k+2)} = \frac{1}{2k} - \frac{1}{k+1} + \frac{1}{2(k+2)...
The sum becomes:
\sum_{k=1}^{50} \left( \frac{1}{2k} - \frac{1}{k+1} + \frac{1}{2(k+2)}...
Reindex the sums:
= \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} \right) - \left( \sum...
Combine terms:

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\hline
& 1 & -5 & 6 & 0 \
\end{array}
The quotient is \( x^2 - 5x + 6 \), so:
\[ x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6) \]
Factor \( x^2 - 5x + 6 \):
\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]
Thus, the polynomial factors as:
\[ (x + 1)(x - 2)(x - 3) \]
We solve \( (x + 1)(x - 2)(x - 3) < 0 \) by analyzing the sign changes...

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