Exercise 2Calculus › Single-variable calculus › SequencesHigh schoolAdded 2 Nov 2026

A recurrence and a proof by induction

#induction#recurrence#closed form

Statement

Let $u_0=1$ and $u_{n+1}=u_n+n+2$ .

Compute $u_1,u_2,u_3$ . Is $(u_n)$ arithmetic?
Prove by induction that $u_n=\dfrac{(n+1)(n+2)}2$ for all $n\ge0$ .

Hint

For part 1, look at the differences $u_{n+1}-u_n$ . For part 2, check the base case, then plug the formula into the recurrence.

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Solution

1. $u_1=3$ , $u_2=6$ , $u_3=10$ . The differences are $2,3,4,\dots$ , not constant, so the sequence is not arithmetic.

2. Base case. $u_0=1=\frac{1\cdot2}2$ .

Inductive step. Assume $u_n=\frac{(n+1)(n+2)}2$ . Then
$\begin{aligned} u_{n+1}&=\frac{(n+1)(n+2)}2+(n+2) \\ &=\frac{(n+2)\bigl[(n+1)+2\bigr]}2 \\ &=\frac{(n+2)(n+3)}2 \end{aligned}$
which is the formula at rank $n+1$ .

Conclusion. We conclude $u_n=\frac{(n+1)(n+2)}2$ .

A recurrence and a proof by induction

Statement

Solution

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