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1.1.3 Evaluate \( \sum_{P=1}^{n} T_{P} \)

Ask by Summers Moran. in South Africa
Mar 16,2025

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The sum \( \sum_{P=1}^{n} T_P \) equals \( \frac{n(n+1)(n+2)}{6} \).

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To evaluate \( \sum_{P=1}^{n} T_{P} \), we first need to identify what \( T_{P} \) represents. If \( T_{P} \) is a specific term or sequence defined in the context of the problem, such as triangular numbers or another well-known sequence, the evaluation will depend on that definition. For example, if \( T_{P} \) represents the \( P \)-th triangular number, then \( T_{P} = \frac{P(P+1)}{2} \). The summation can then be expressed as: \[ \sum_{P=1}^{n} T_{P} = \sum_{P=1}^{n} \frac{P(P+1)}{2} = \frac{1}{2} \sum_{P=1}^{n} P(P+1) \] Using the formula for the sum of the first \( n \) integers, \( \sum_{P=1}^{n} P = \frac{n(n+1)}{2} \), and the formula for the sum of squares \( \sum_{P=1}^{n} P^2 = \frac{n(n+1)(2n+1)}{6} \), we can compute this further. By substituting these values, we can arrive at a closed formula for \( \sum_{P=1}^{n} T_{P} \). For different definitions of \( T_{P} \), the process would adjust accordingly but follow a similar reasoning. Ultimately, the evaluation is contingent upon defining \( T_{P} \) properly within its given context.

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