Prove that 1+2+3++n = sum_k=1^n k = (n(n+1))/(2), n 1.

March 29, 2026Mathematics

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March 29, 2026

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\sum_{k=1}^{n} k = \frac{n(n+1)}{2} $,$ \forall n \ge 1$.

Here are the proofs using the Principle of Mathematical Induction.

1. Prove that $1+2+3+\dots+n = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ , $\forall n \ge 1$ .

Let $P(n)$ be the statement $1+2+3+\dots+n = \frac{n(n+1)}{2}$ .

Base Case (n=1): LHS $= 1$ RHS $= \frac{1(1+1)}{2} = \frac{1(2)}{2} = 1$ Since LHS = RHS, $P(1)$ is true.
Inductive Hypothesis: Assume $P(m)$ is true for some positive integer $m$ . That is, assume $1+2+3+\dots+m = \frac{m(m+1)}{2}$ .
Inductive Step: We need to prove that $P(m+1)$ is true.

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LHS = 1 RHS = (1(1+1))/(2) = (1(2))/(2) = 1 Since LHS = RHS, P(1) is true. Inductive Hypothesis: Assume P(m) is true for some positive integer m.

Asked on March 29, 2026Mathematics

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ScanSolve AI AnswerAnswered on March 29, 2026

Here are the proofs using the Principle of Mathematical Induction. 1. Prove that 1+2+3++n = _k=1^n k = (n(n+1))/(2), n 1. Let P(n) be the statement 1+2+3++n = (n(n+1))/(2). Base Case (n=1): LHS = 1 RHS = (1(1+1))/(2) = (1(2))/(2) = 1 Since LHS = RHS, P(1) is true. Inductive Hypothesis: Assume P(m) is true for some positive integer m. That is, assume 1+2+3++m = (m(m+1))/(2). Inductive Step: We need to prove that P(m+1) is true.