An integer upper limit in an inequality · source reference

Printed question

Find the set of values of $a$ for which $\sum_{r=3}^{a}(2r+a)>a^2$. [5]

Printed working

$$\begin{aligned}\sum_{r=3}^a(2r+a)&>a^2\\\frac{a-3+1}2[(a+6)+(2a+a)]&>a^2\\\frac{a-2}2(4a+6)-a^2&>0\\2a^2-a-6-a^2&>0\\a^2-a-6&>0\\(a-3)(a+2)&>0.\end{aligned}$$

$a>3$ or $a<-2$ (reject because upper limit $a\ge3$). Required set: $\{a:a>3,a\in\mathbb Z^+\}$.

Alternatively,

$$\begin{aligned}2\sum_{r=3}^ar+\sum_{r=3}^aa&>a^2\\2\left(\sum_{r=1}^ar-\sum_{r=1}^2r\right)+(a-3+1)a&>a^2\\2\left[\frac a2(1+a)-(1+2)\right]+(a^2-2a)-a^2&>0\\a^2+a-6-2a&>0\\a^2-a-6&>0\\(a-3)(a+2)&>0.\end{aligned}$$

Again $a>3$ or $a<-2$ (rejected), giving $\{a:a>3,a\in\mathbb Z^+\}$.

Teacher notes

LHS is an AP sum:

$$\sum_{r=3}^a(2r+a)=(6+a)+(8+a)+\cdots+(2a+a).$$

First term $6+a$; common difference 2; number of terms $(a-3)+1=(\text{upper limit}-\text{lower limit})+1$. Note that $r$ is a variable while $a$ is a constant. Upper limit must be at least the lower limit, here 3.

Source diagrams redrawn

Printed answer key

$\{a:a>3,a\in\mathbb Z^+\}$

Mathematical review and qualifications

For integer $a\ge3$, the sum is $(a-2)(4a+6)/2=2a^2-a-6$. The inequality becomes $(a-3)(a+2)>0$, giving $a<-2$ or $a>3$. Only the latter respects the sum bounds. Thus $a\in\mathbb Z$, $a\ge4$.

Do not keep the algebraic branch a<−2: it is incompatible with the stated finite-sum indexing.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [2015/TJC/MCT/1] Question PDF pages 1; solution PDF pages 1. The original files remain private QA records outside this native content.