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Home » 27a + 9b + 3c + d = 2 - Dyverse

27a + 9b + 3c + d = 2 - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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Multiply both sides by $ k(k+2) $:
= A(k+2) + Bk.
Set $ k = 0 $: $ 1 = A(2) \Rightarrow A = rac{1}{2} $.
Set $ k = -2 $: $ 1 = B(-2) \Rightarrow B = -rac{1}{2} $.
rac{1}{k(k+2)} = rac{1}{2}\left( rac{1}{k} - rac{1}{k+2}
Now substitute into the sum:
\sum_{k=1}^{50} rac{1}{k(k+2)} = rac{1}{2} \sum_{k=1}^{50} \left( r...
This is a telescoping series. Write out the terms:
rac{1}{2} \left[ \left( rac{1}{1} - rac{1}{3}
ight) + \left( rac{1}{2} - rac{1}{4}

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