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Home » \frac1k(k+2) = \fracAk + \fracBk+2 - Network4

\frac1k(k+2) = \fracAk + \fracBk+2 - Network4

📅 May 19, 2026 👤 scraface

May 19, 2026

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-\frac{7}{6} + \frac{21}{2} - \frac{82}{3} + d = 3
-\frac{7}{6} + \frac{63}{6} - \frac{164}{6} + d = 3
-\frac{108}{6} + d = 3 \implies -18 + d = 3 \implies d = 21
Thus, the polynomial is:
f(x) = -\frac{7}{6}x^3 + \frac{21}{2}x^2 - \frac{82}{3}x + 21
\boxed{-\frac{7}{6}x^3 + \frac{21}{2}x^2 - \frac{82}{3}x + 21}
Let \( a \) and \( b \) be complex numbers such that \( a + b = 4 + 2i...
We start with the identity for the square of a sum:
Rearranging gives:
a^2 + b^2 = (a + b)^2 - 2ab

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