A polynomial term and a middle block · source reference

Printed question

Given $\sum_{r=1}^nr^2=\frac n6(n+1)(2n+1)$, show $\sum_{r=1}^n(2r+1)(2r+3)=\frac n3(4n^2+18n+23)$. Hence find the exact sum of the 25th to the 75th term of the series. [5]

Printed working

$$\begin{aligned}\sum_{r=1}^n(2r+1)(2r+3)&=\sum_{r=1}^n(4r^2+8r+3)\\&=4\sum_{r=1}^nr^2+\sum_{r=1}^n(8r+3)\\&=4\left[\frac n6(n+1)(2n+1)\right]+\frac n2(11+8n+3)\\&=\frac{2n}3(n+1)(2n+1)+\frac n2(8n+14)\\&=\frac n3(4n^2+18n+23).\end{aligned}$$ $$\begin{aligned}\sum_{r=25}^{75}(2r+1)(2r+3)&=\sum_{r=1}^{75}(2r+1)(2r+3)-\sum_{r=1}^{24}(2r+1)(2r+3)\\&=\frac{75}3[4(75)^2+18(75)+23]-\frac{24}3[4(24)^2+18(24)+23]\\&=596825-22072=574753.\end{aligned}$$

Teacher notes

$\sum_{r=1}^n(8r+3)=11+19+26+\cdots+(8n+3)$ as printed. It is an AP with first term 11, common difference 8, $n$ terms, $S_n=\frac n2(u_1+u_n)$.

The displayed “26” is a source typo: the third term is $8(3)+3=27$. The working correctly uses common difference 8.

To use the result, the lower limit of $r$ must be 1.

Printed answer key

574753

Mathematical review and qualifications

$S_n=4[n(n+1)(2n+1)/6]+8[n(n+1)/2]+3n=n(4n^2+18n+23)/3$. The required inclusive block is $S_{75}-S_{24}=596825-22072=574753$.

The lower endpoint is included: subtract S_24, not S_25.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [2023/DHS/Prelim/II/2] Question PDF pages 1; solution PDF pages 3. The original files remain private QA records outside this native content.