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