A radical sum with one extra endpoint · source reference

Printed question

Given $\sum_{r=1}^{2500}\frac1{\sqrt{4r-3}+\sqrt{4r+1}}=\frac{\sqrt{10001}-1}4$, hence show, without a calculator, that $\sum_{r=0}^{2500}\frac1{\sqrt{4r+1}+\sqrt{4r+5}}>24$. [3]

Printed working

Replace $r$ by $r-1$:

$$\begin{aligned}\sum_{r=0}^{2500}\frac1{\sqrt{4r+1}+\sqrt{4r+5}}&=\sum_{r=1}^{2501}\frac1{\sqrt{4r-3}+\sqrt{4r+1}}\\&=\sum_{r=1}^{2500}\frac1{\sqrt{4r-3}+\sqrt{4r+1}}+\frac1{\sqrt{10001}+\sqrt{10005}}\\&=\frac{\sqrt{10001}-1}4+\frac1{\sqrt{10001}+\sqrt{10005}}\\&>\frac{\sqrt{10000}-1}4+\frac1{\sqrt{10001}+\sqrt{10005}}\\&=24.75+\frac1{\sqrt{10001}+\sqrt{10005}}>24.\end{aligned}$$

Teacher notes

Remember to change $r$ to $r-1$ for the upper limit too. $\sum_{r=1}^{2501}T_r=\sum_{r=1}^{2500}T_r+T_{2501}$.

Source typographic slip: its penultimate denominator reads $\sqrt{10001}+\sqrt{10003}$ rather than the earlier correct $\sqrt{10001}+\sqrt{10005}$. Both are positive, but the equality requires 10005; that correction is separate from the printed record.

Printed answer key

No separate short-answer entry is printed for this question; the worked solution is retained above.

Mathematical review and qualifications

After shifting, the sum is $(\sqrt{10001}-1)/4+1/(\sqrt{10001}+\sqrt{10005})$. Since $10001>10000$, the first term exceeds $(100-1)/4=99/4=24.75$, and the extra term is positive. Thus the sum exceeds 24 without numerical evaluation.

There are 2501 terms in the requested sum because its lower index is zero.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: SIGMA NOTATION. [modified 2012/ACJC/Prelim/I/10] Question PDF pages 2; solution PDF pages 9. The original files remain private QA records outside this native content.