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📅 May 5, 2026 👤 scraface

May 05, 2026

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\frac{1}{k(k+1)(k+2)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2}.
Multiply through: $ 1 = A(k+1)(k+2) + Bk(k+2) + Ck(k+1) $.
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:

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