Sum a block of odd squares · source reference

Printed question

It is given that $\sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}6$.

(i) Show $\sum_{r=1}^n(2r-1)^2=\frac{n(4n^2-1)}3$. [3]

(ii) Hence find an expression for $13^2+15^2+\cdots+(4m-1)^2$ in terms of $m$. [4]

Printed working

(i)

$$\begin{aligned}\sum_{r=1}^n(2r-1)^2&=\sum_{r=1}^n(4r^2-4r+1)\\&=4\sum_{r=1}^nr^2-4\sum_{r=1}^nr+\sum_{r=1}^n1\\&=4\left[\frac{n(n+1)(2n+1)}6\right]-4\left[\frac{n(n+1)}2\right]+n\\&=\frac{2n(n+1)(2n+1)}3-2n(n+1)+n\\&=\frac{2n(n+1)(2n+1)-6n(n+1)+3n}3\\&=\frac{n[2(n+1)(2n+1)-6(n+1)+3]}3\\&=\frac{n(4n^2+6n+2-6n-6+3)}3=\frac{n(4n^2-1)}3.\end{aligned}$$

Alternative method: $4\sum_{r=1}^nr^2+\sum_{r=1}^n(-4r+1)$,

$$4\left[\frac{n(n+1)(2n+1)}6\right]+\frac n2[-3+(-4n+1)].$$

This simplifies to the same result.

(ii) $2r-1=13\Rightarrow r=7$; $2r-1=4m-1\Rightarrow r=2m$.

$$\begin{aligned}13^2+15^2+\cdots+(4m-1)^2&=\sum_{r=7}^{2m}(2r-1)^2\\&=\sum_{r=1}^{2m}(2r-1)^2-\sum_{r=1}^6(2r-1)^2\\&=\frac{2m[4(2m)^2-1]}3-\frac{6[4(6)^2-1]}3\\&=\frac{2m(16m^2-1)}3-286.\end{aligned}$$

Teacher notes

EXPAND! Use the sum of an AP with $a=1,d=1,n$ terms for $\sum r$, and $n$ copies for $\sum1$.

Alternative: $\sum_{r=1}^n(-4r+1)=-3-7-11-\cdots-(4n-1)$ is an AP with first term −3, common difference −4, $n$ terms; $S_n=\frac n2(u_1+u_n)$.

Identify the lower and upper limits. To use (i), the lower limit of $r$ must be 1.

Printed answer key

(ii) $\frac{2m(16m^2-1)}3-286$

Mathematical review and qualifications

(i) $\sum(2r-1)^2=4\sum r^2-4\sum r+n=\frac23n(n+1)(2n+1)-2n(n+1)+n=n(4n^2-1)/3$.

(ii) The block runs from $r=7$ to $r=2m$, so its sum is $S_{2m}-S_6=\frac{2m(16m^2-1)}3-286$ for integer $m\ge4$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [2024/MI/Prelim/I/9b] Question PDF pages 1; solution PDF pages 1, 2. The original files remain private QA records outside this native content.