Exercise 252Logic, sets and proofs › Proof techniques › InductionHigh schoolAdded 3 Nov 2026

Proof by induction: sum of first $n$ integers

#induction#sum#integers

Statement

Prove by induction that $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ for all $n\ge1$ .

Hint

Base case $n=1$ ; inductive step: add $n+1$ to both sides.

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Solution

Base case. $n=1$ : $\sum_{k=1}^1k=1=\frac{1\cdot2}{2}$ . ✓

Inductive step. Assume $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ . Then
$\sum_{k=1}^{n+1}k=\frac{n(n+1)}{2}+(n+1)=\frac{n(n+1)+2(n+1)}{2}=\frac{(n+1)(n+2)}{2}.$

Conclusion. By induction, the formula holds for all $n\ge1$ .

Proof by induction: sum of first $n$ integers

Statement

Solution

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