\sum_{k=1}^{50} \left( \frac{1}{2k} - \frac{1}{k+1} + \frac{1}{2(k+2)} \right) = \frac{1}{2} \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=1}^{50} \frac{1}{k+1} + \frac{1}{2} \sum_{k=1}^{50} \frac{1}{k+2}.

["Title: Simplify the Summation: A Step-by-Step Proof Using Algebra and Series Properties", "---", "Summations can often appear complex at first glance, especially when dealing with nested fractions and alternating signs. A powerful technique in mathematics involves telescoping and reindexing to simplify expressions. This article explains how to prove the equality:", "[\n\sum_{k=1}^{50} \left( \frac{1}{2k} - \frac{1}{k+1} + \frac{1}{2(k+2)} \right) = \frac{1}{2} \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=1}^{50} \frac{1}{k+1} + \frac{1}{2} \sum_{k=1}^{50} \frac{1}{k+2}\n]", "using algebraic manipulation and series manipulation.", "---", "### Break Down the Left-Hand Side (LHS)", "We start with the original left-hand side:", "[\n\sum_{k=1}^{50} \left( \frac{1}{2k} - \frac{1}{k+1} + \frac{1}{2(k+2)} \right)\n]", "Observe that each term in the summation consists of three fractions. We aim to separate this sum into three independent summations:", "[\n= \sum_{k=1}^{50} \frac{1}{2k} - \sum_{k=1}^{50} \frac{1}{k+1} + \sum_{k=1}^{50} \frac{1}{2(k+2)}\n]", "Now rename the indices in the last sum to enable reindexing.", "---", "### Reindex the Sums for Uniformity", "First term:\n[\n\sum_{k=1}^{50} \frac{1}{2k} = \frac{1}{2} \sum_{k=1}^{50} \frac{1}{k}\n]", "Second term:\n[\n\sum_{k=1}^{50} \frac{1}{k+1} = \sum_{j=2}^{51} \frac{1}{j} \quad \ ext{(let } j = k+1\ ext{)}\n]", "Third term:\n[\n\sum_{k=1}^{50} \frac{1}{2(k+2)} = \frac{1}{2} \sum_{k=1}^{50} \frac{1}{k+2} = \frac{1}{2} \sum_{m=3}^{52} \frac{1}{m} \quad \ ext{(let } m = k+2\ ext{)}\n]", "---", "### Rewrite the Entire Expression", "Now substitute back:", "[\n\ ext{LHS} = \frac{1}{2} \sum_{k=1}^{50} \frac{1}{k} - \sum_{j=2}^{51} \frac{1}{j} + \frac{1}{2} \sum_{m=3}^{52} \frac{1}{m}\n]", "Now replace dummy indices consistently (we can rewrite everything in terms of ( k )):", "[\n\ ext{LHS} = \frac{1}{2} \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=2}^{51} \frac{1}{k} + \frac{1}{2} \sum_{k=3}^{52} \frac{1}{k}\n]", "---", "### Express All Sums in Terms of Full Harmonic Sums", "We write each summation from ( k=1 ) to ( 52 ), adjusting for missing terms:", "- (\sum_{k=1}^{50} \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \sum_{k=3}^{50} \frac{1}{k})\n- (\sum_{k=2}^{51} \frac{1}{k} = \frac{1}{2} + \sum_{k=3}^{50} \frac{1}{k} + \frac{1}{51})\n- (\sum_{k=3}^{52} \frac{1}{k} = \sum_{k=3}^{50} \frac{1}{k} + \frac{1}{51} + \frac{1}{52})", "Substitute into LHS:", "[\n\ ext{LHS} = \frac{1}{2} \left( 1 + \frac{1}{2} + \sum_{k=3}^{50} \frac{1}{k} \right) - \left( \frac{1}{2} + \sum_{k=3}^{50} \frac{1}{k} + \frac{1}{51} \right) + \frac{1}{2} \left( \sum_{k=3}^{50} \frac{1}{k} + \frac{1}{51} + \frac{1}{52} \right)\n]", "---", "### Expand and Combine Like Terms", "Expand each part:", "- First term:\n[\n\frac{1}{2} + \frac{1}{4} + \frac{1}{2} \sum_{k=3}^{50} \frac{1}{k}\n]", "- Second term:\n[\n- \frac{1}{2} - \sum_{k=3}^{50} \frac{1}{k} - \frac{1}{51}\n]", "- Third term:\n[\n+ \frac{1}{2} \sum_{k=3}^{50} \frac{1}{k} + \frac{1}{2 \cdot 51} + \frac{1}{2 \cdot 52}\n]", "Now combine constant terms and harmonic sums:", "- Constants:\n[\n\left( \frac{1}{2} + \frac{1}{4} - \frac{1}{2} \right) = \frac{1}{4}, \quad -\frac{1}{51} + \frac{1}{2 \cdot 51} = -\frac{1}{102}\n]", "- Harmonic sums:\n[\n\frac{1}{2} \sum_{k=3}^{50} \frac{1}{k} - \sum_{k=3}^{50} \frac{1}{k} + \frac{1}{2} \sum_{k=3}^{50} \frac{1}{k} = \left( \frac{1}{2} - 1 + \frac{1}{2} \right) \sum_{k=3}^{50} \frac{1}{k} = 0\n]", "Now combine the remaining constants:", "[\n\frac{1}{4} - \frac{1}{102} + \frac{1}{102} = \frac{1}{4}\n]", "Wait — this suggests cancellation too strongly, so double-check.", "Actually, we missed: (-\frac{1}{51}) and (+\frac{1}{2 \cdot 51} = -\frac{1}{51} + \frac{1}{102} = -\frac{2}{102} + \frac{1}{102} = -\frac{1}{102}), and (+\frac{1}{2 \cdot 52} = \frac{1}{104})", "But also in the first sum, there is (-\frac{1}{51}) and (+\frac{1}{4} - \frac{1}{2}) — correction again.", "Let's re-collect all constant parts:", "- From first term: ( \frac{1}{2} \cdot 1 = \frac{1}{2} ), ( \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} ), rest from (k \geq 3:\frac{1}{2} \sum)\n- From second term: ( -\frac{1}{2} ), ( -\frac{1}{51} )\n- From third term: ( +\frac{1}{2 \cdot 51} = -\frac{1}{102} ), ( +\frac{1}{2 \cdot 52} = +\frac{1}{104} )", "Now constants alone:", "[\n\frac{1}{2} + \frac{1}{4} - \frac{1}{2} - \frac{1}{51} + \frac{1}{102} + \frac{1}{104}\n]", "Simplify:", "[\n\left( \frac{1}{2} - \frac{1}{2} \right) + \frac{1}{4} - \frac{1}{51} + \frac{1}{102} + \frac{1}{104} = \frac{1}{4} - \frac{2}{102} + \frac{1}{102} + \frac{1}{104} = \frac{1}{4} - \frac{1}{102} + \frac{1}{104}\n]", "But this is messy — something is off.", "Better approach: keep the expression in pooled summations and isolate telescoping behavior directly.", "---", "### Alternative: Use Telescoping by Expansion", "Return to the inner expression per ( k ):", "[\n\frac{1}{2k} - \frac{1}{k+1} + \frac{1}{2(k+2)}\n]", "We aim to write it as a difference of harmonic-like terms.", "Observe that:", "[\n\frac{1}{2k} - \frac{1}{k+1} + \frac{1}{2(k+2)} = \frac{1}{2} \left( \frac{1}{k} - \frac{2}{k+1} + \frac{1}{k+2} \right)\n]", "Now, consider the telescoping nature of:", "[\n\frac{1}{2} \left( \frac{1}{k} - \frac{2}{k+1} + \frac{1}{k+2} \right)\n= \frac{1}{2} \left( \left( \frac{1}{k} - \frac{1}{k+1} \right) - \left( \frac{1}{k+1} - \frac{1}{k+2} \right) \right)\n]", "Check:", "[\n\left( \frac{1}{k} - \frac{1}{k+1} \right) - \left( \frac{1}{k+1} - \frac{1}{k+2} \right) = \frac{1}{k} - \frac{2}{k+1} + \frac{1}{k+2}\n]", "Exactly!", "So:", "[\n\frac{1}{2k} - \frac{1}{k+1} + \frac{1}{2(k+2)} = \frac{1}{2} \left( \left( \frac{1}{k} - \frac{1}{k+1} \right) - \left( \frac{1}{k+1} - \frac{1}{k+2} \right) \right)\n]", "Now sum from ( k=1 ) to ( 50 ):", "[\n\sum_{k=1}^{50} \left( \cdots \right) = \frac{1}{2} \sum_{k=1}^{50} \left[ \left( \frac{1}{k} - \frac{1}{k+1} \right) - \left( \frac{1}{k+1} - \frac{1}{k+2} \right) \right]\n]", "The sum becomes:", "[\n= \frac{1}{2} \left( \sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+1} \right) - \sum_{k=1}^{50} \left( \frac{1}{k+1} - \frac{1}{k+2} \right) \right)\n]", "Each sum is telescoping.", "First:\n[\n\sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{50} - \frac{1}{51}\right) = 1 - \frac{1}{51}\n]", "Second:\n[\n\sum_{k=1}^{50} \left( \frac{1}{k+1} - \frac{1}{k+2} \right) = \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{51} - \frac{1}{52}\right) = \frac{1}{2} - \frac{1}{52}\n]", "Subtract:", "[\n\left(1 - \frac{1}{51}\right) - \left(\frac{1}{2} - \frac{1}{52}\right) = 1 - \frac{1}{51} - \frac{1}{2} + \frac{1}{52} = \frac{1}{2} - \frac{1}{51} + \frac{1}{52}\n]", "Multiply by ( \frac{1}{2} ):", "[\n\ ext{RHS} = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{51} + \frac{1}{52} \right) = \frac{1}{4} - \frac{1}{2 \cdot 51} + \frac{1}{2 \cdot 52}\n]", "But we now compare with the LHS original expression:", "We previously expressed:", "[\n\ ext{LHS} = \frac{1}{2} \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=1}^{50} \frac{1}{k+1} + \frac{1}{2} \sum_{k=1}^{50} \frac{1}{k+2}\n]", "Rewriting indices and combining:", "Let’s compute RHS explicitly:", "[\n\frac{1}{2} \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=2}^{51} \frac{1}{k} + \frac{1}{2} \sum_{k=3}^{52} \frac{1}{k}\n]", "Now write all in terms of ( \sum_{k=1}^{52} ) with coefficients:", "- ( \frac{1}{2} \left( \frac{1}{1} + \frac{1}{2} + \sum_{3}^{50} \frac{1}{k} \right) = \frac{1}{2} + \frac{1}{4} + \frac{1}{2} \sum_{3}^{50} \frac{1}{k} )\n- ( - \left( \frac{1}{2} + \sum_{3}^{50} \frac{1}{k} + \frac{1}{51} \right) )\n- ( + \frac{1}{2} \left( \sum_{3}^{50} \frac{1}{k} + \frac{1}{51} + \frac{1}{52} \right) )", "Sum coefficient by coefficient:", "- ( k=1 ): ( \frac{1}{2} )\n- ( k=2 ): ( \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} )\n- ( k=3 ) to ( 50 ): ( \frac{1}{2} - 1 + \frac{1}{2} = 0 )\n- ( k=51 ): ( - \frac{1}{51} + \frac{1}{2} \cdot \frac{1}{51} = -\frac{1}{102} )\n- ( k=52 ): ( \frac{1}{2} \cdot \frac{1}{52} = \frac{1}{104} )", "So total:", "[\n\frac{1}{2} - \frac{1}{4} - \frac{1}{102} + \frac{1}{104} = \frac{1}{4} - \frac{1}{102} + \frac{1}{104}\n]", "Now compute numerically or simplify algebraically:", "But earlier transformed version gave:", "[\n\ ext"]