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H1 Math

H2 Math

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IB AA SL Math

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IB AI SL Math

IP Math

N(A) A Math

N(A) E Math

PSLE

PSLE Foundation Mathematics

<p>The sum, \(S_n\), of the first \(n\) terms of a sequence \(u_1,u_2,u_3,\ldots\) is given by \[S_n=\frac16+\frac1{n+3}-\frac2{n+4}.\]</p>
<ol class="abc-list"><li data-list-item-id="nyjc-q05-item01" data-list-item-path="nyjc-q05-item01"><div class="list-item-paragraph">Give a reason why the series \(\displaystyle \sum_{r=1}^{\infty}u_r\) converges and write down the value of the sum to infinity.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="nyjc-q05-item02" data-list-item-path="nyjc-q05-item02"><div class="list-item-paragraph">Show that&nbsp;\(\displaystyle u_n=\frac{n}{(n+2)(n+3)(n+4)}\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="nyjc-q05-item03" data-list-item-path="nyjc-q05-item03"><div class="list-item-paragraph">It is given that&nbsp;\(\displaystyle 5\sum_{r=k+1}^{\infty}u_r=9\sum_{r=1}^{k}u_r\). Find the value of \(k\).<span class="inline-marks">[3]</span></div></li></ol>

The sum, \(S_n\), of the first \(n\) terms of a sequence \(u_1,u_2,u_3,\ldots\) is given by \[S_n=\frac16+\frac1{n+3}-\frac2{n+4}.\]

Give a reason why the series \(\displaystyle \sum_{r=1}^{\infty}u_r\) converges and write down the value of the sum to infinity.

Show that \(\displaystyle u_n=\frac{n}{(n+2)(n+3)(n+4)}\).

It is given that \(\displaystyle 5\sum_{r=k+1}^{\infty}u_r=9\sum_{r=1}^{k}u_r\). Find the value of \(k\).

Manipulating Summation Limits

Application of Sum to Infinity

\(u_n = S_n - S_{n-1} \)

2026 NYJC P1 Q5

Sequence and Series

<ol class="abc-list"><li data-list-item-id="asrjc-q10-item01" data-list-item-path="asrjc-q10-item01"><div class="list-item-paragraph">Given that&nbsp;\(k\) is a real constant, find \(\displaystyle \sum_{r=5}^{30}[2^{1-2r}-(k-2r)]\), leaving your answer in the form \(A(1-B)+Ck+910\), where&nbsp;\(A\), \(B\) and \(C\) are exact real constants.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="asrjc-q10-item02" data-list-item-path="asrjc-q10-item02"><div class="list-item-paragraph">Another series is such that \[\sum_{r=1}^{N}\frac1{r(r+1)(r+2)}=\frac12\left(\frac12-\frac1{(N+1)(N+2)}\right).\]</div><ol class="i-list"><li data-list-item-id="asrjc-q10-item02-sub-item01" data-list-item-path="asrjc-q10-item02/asrjc-q10-item02-sub-item01"><div class="list-item-paragraph">Find \(\displaystyle \sum_{r=1}^{\infty}\frac1{r(r+1)(r+2)}\). Explain your answer clearly.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="asrjc-q10-item02-sub-item02" data-list-item-path="asrjc-q10-item02/asrjc-q10-item02-sub-item02"><div class="list-item-paragraph">Find \(\displaystyle \sum_{r=13}^{N}\frac1{3r(r-1)(r-2)}\) in terms of \(N\).<span class="inline-marks">[3]</span></div></li></ol></li><li data-list-item-id="asrjc-q10-item03" data-list-item-path="asrjc-q10-item03"><div class="list-item-paragraph">A sequence of positive numbers \(x_1,x_2,x_3,\ldots\) satisfies the recurrence relation \[x_{n+1}=\sqrt{10-3x_n},\qquad n=1,2,3,\ldots\]</div><ol class="i-list"><li data-list-item-id="asrjc-q10-item03-sub-item01" data-list-item-path="asrjc-q10-item03/asrjc-q10-item03-sub-item01"><div class="list-item-paragraph">Given that&nbsp;\(x_4=1\), find the exact value of \(x_1\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="asrjc-q10-item03-sub-item02" data-list-item-path="asrjc-q10-item03/asrjc-q10-item03-sub-item02"><div class="list-item-paragraph">It is also known that the sequence converges to \(l\). Find the value of \(l\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="asrjc-q10-item03-sub-item03" data-list-item-path="asrjc-q10-item03/asrjc-q10-item03-sub-item03"><div class="list-item-paragraph">Prove that&nbsp;\((x_{n+1})^2-l^2=3(l-x_n)\).<span class="inline-marks">[1]</span></div></li><li data-list-item-id="asrjc-q10-item03-sub-item04" data-list-item-path="asrjc-q10-item03/asrjc-q10-item03-sub-item04"><div class="list-item-paragraph"><p>Use the result in part <strong>(iii)</strong> to show that if \(x_n&lt;l\), then \(x_{n+1}&gt;l \).</p><span class="inline-marks">[2]</span></div></li></ol></li></ol>

Given that \(k\) is a real constant, find \(\displaystyle \sum_{r=5}^{30}[2^{1-2r}-(k-2r)]\), leaving your answer in the form \(A(1-B)+Ck+910\), where \(A\), \(B\) and \(C\) are exact real constants.

Sigma Notation By Substitution

Limit of a Sequence and Behaviour of a Sequence

Reverse Iteration of Recurrence Relations

Summations Involving APGP

2026 ASRJC P1 Q10

<ol class="abc-list"><li data-list-item-id="h2-2025-yijc-promo-q4-yijc-q4-item-block-0-item-0" data-list-item-path="h2-2025-yijc-promo-q4-yijc-q4-item-block-0-item-0"><div class="list-item-paragraph">Given that&nbsp;\(\sum_{r=1}^{n} r^2=\frac{n}{6}(n+1)(2n+1)\), find \(\sum_{r=1}^{n}(4r+1)(6r+1)\) in terms of \(n\).<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-yijc-promo-q4-yijc-q4-item-block-0-item-1" data-list-item-path="h2-2025-yijc-promo-q4-yijc-q4-item-block-0-item-1"><div class="list-item-paragraph">Hence find \(\sum_{r=8}^{2n}(4r+1)(6r+1)\) in terms of \(n\).<span class="inline-marks">[3]</span></div></li></ol>

Given that \(\sum_{r=1}^{n} r^2=\frac{n}{6}(n+1)(2n+1)\), find \(\sum_{r=1}^{n}(4r+1)(6r+1)\) in terms of \(n\).

Hence find \(\sum_{r=8}^{2n}(4r+1)(6r+1)\) in terms of \(n\).

Summing Polynomial Series

2025 YIJC Promo Q4

<p>It is given that&nbsp;\( \sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4} \).</p>
<ol class="abc-list"><li data-list-item-id="h2-2025-vjc-promo-q3-vjc-q3-item-block-1-item-0" data-list-item-path="h2-2025-vjc-promo-q3-vjc-q3-item-block-1-item-0"><div class="list-item-paragraph">Find \( \sum_{r=4}^{N+3} (r-2)^3 \) in terms of \(N\).<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-vjc-promo-q3-vjc-q3-item-block-1-item-1" data-list-item-path="h2-2025-vjc-promo-q3-vjc-q3-item-block-1-item-1"><div class="list-item-paragraph">Hence, find the largest value of \(N\) for which \( \sum_{r=4}^{N+3} \left[(2r-4)^3+2^r\right] &lt; 10^5 \).<span class="inline-marks">[3]</span></div></li></ol>

It is given that \( \sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4} \).

Find \( \sum_{r=4}^{N+3} (r-2)^3 \) in terms of \(N\).

Hence, find the largest value of \(N\) for which \( \sum_{r=4}^{N+3} \left[(2r-4)^3+2^r\right] < 10^5 \).

Problems involving inequalities

Geometric Progression

2025 VJC Promo Q3

<p>Given that&nbsp;\(\displaystyle \sum_{r=1}^{n}\frac{2}{(r+4)(r+6)}=\frac{11}{30}-\frac{1}{n+5}-\frac{1}{n+6}\),</p>
<ol class="i-list"><li data-list-item-id="q2-item-i" data-list-item-path="q2-item-i"><div class="list-item-paragraph">Find \(\displaystyle \sum_{r=5}^{n+4}\frac{2}{r(r+2)}\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="q2-item-ii" data-list-item-path="q2-item-ii"><div class="list-item-paragraph">Hence, show that&nbsp;\(\displaystyle \sum_{r=5}^{n+4}\frac{1}{(r+1)^2}&lt;\frac{11}{60}\).<span class="inline-marks">[3]</span></div></li></ol>

Given that \(\displaystyle \sum_{r=1}^{n}\frac{2}{(r+4)(r+6)}=\frac{11}{30}-\frac{1}{n+5}-\frac{1}{n+6}\),

Reindexed rational sum and an upper bound

<p>Given that&nbsp;\(\displaystyle \sum_{r=1}^{n}\frac{1}{(3r-1)(3r+2)}=\frac{1}{6}-\frac{1}{3(3n+2)}\), find \(\displaystyle \sum_{r=4}^{n}\frac{1}{(3r+2)(3r+5)}\) in terms of \(n\).<span class="inline-marks">[4]</span></p>

Given that \(\displaystyle \sum_{r=1}^{n}\frac{1}{(3r-1)(3r+2)}=\frac{1}{6}-\frac{1}{3(3n+2)}\), find \(\displaystyle \sum_{r=4}^{n}\frac{1}{(3r+2)(3r+5)}\) in terms of \(n\).

Shifted rational sum with omitted terms

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">For this question, you may use these results:</div><div class="list-item-paragraph" style="text-align: center">\[\sum_{r=1}^{n}r^2 = \frac{n(n+1)(2n+1)}{6} \quad \text{and} \quad \sum_{r=1}^{n}r^3 = \frac{n^2 (n+1)^2}{4}.\]</div><ol class="i-list"><li data-list-item-id="item-a-i" data-list-item-path="item-a/item-a-i"><div class="list-item-paragraph">Show that&nbsp;\(\sum_{r=1}^{n}r(r+1)^2 = \frac{n(n+1)(n+2)(3n+5)}{12}\).<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-a-ii" data-list-item-path="item-a/item-a-ii"><div class="list-item-paragraph">Hence find \(\sum_{r=5}^{n-1}(r+2)(r+3)^2\) in terms of \(n\).<span class="inline-marks">[3]</span></div></li></ol></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">The sequence \(u_1, u_2, u_3, \dots\) is defined by \(u_1 = 2\), \(u_{n+1} = \frac{1}{1-u_n}\), \(n \ge 1\).</div><div class="list-item-paragraph">Find the values of \(u_2\), \(u_3\) and \(u_4\). Hence find the value of \(u_{2025}\).<span class="inline-marks">[3]</span></div></li></ol>

Generating Terms from Recurrence Relations

2025 ACJC P2 Q3

<p>It is given that&nbsp;\(\sum_{r=1}^{n} \frac{7r+4}{r(r+1)(r+2)} = \frac{9}{2} - \frac{2}{n+1} - \frac{5}{n+2}\).</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Find \(\sum_{r=7}^{2n} \frac{7r-3}{r^3-r}\) giving your answers in terms of \(n\).<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Show algebraically that&nbsp;\((r+1)^3 &gt; r(r+1)(r+2)\) for all positive integers \(r\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Hence show that&nbsp;\(\sum_{r=1}^{n} \frac{7r+4}{(r+1)^3} &lt; \frac{9}{2}\).<span class="inline-marks">[3]</span></div></li></ol>

It is given that \(\sum_{r=1}^{n} \frac{7r+4}{r(r+1)(r+2)} = \frac{9}{2} - \frac{2}{n+1} - \frac{5}{n+2}\).

2025 ASRJC P1 Q6

<p>Given that&nbsp;\(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\), find an expression, in simplified form, for&nbsp;\(\sum_{r=n+1}^{2n} (2r-1)^2.\)<span class="inline-marks">[6]</span></p>

Given that \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\), find an expression, in simplified form, for \(\sum_{r=n+1}^{2n} (2r-1)^2.\)

N1987 P1 Q19b

<ol class="abc-list"><li data-list-item-id="item-a2" data-list-item-path="item-a2"><div class="list-item-paragraph">The sum, \(S_n\), of the first \(n\) terms of a sequence \(u_1, u_2, u_3, \dots\) is given by \(S_n = 75 - \frac{3^{n+1}}{5^{n-2}}\).<br>Show that&nbsp;\(u_n = k\left(\frac{3}{5}\right)^{n-1}\), where&nbsp;\(k\) is a constant to be determined. Find \(S_\infty\).<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b2" data-list-item-path="item-b2"><div class="list-item-paragraph">Another sequence of numbers \(v_1, v_2, v_3, \dots\) is such that&nbsp;\(\sum_{i=1}^{n} v_i = \frac{2n}{2n + 1}\).<br>Find the exact value of \(\sum_{i=11}^{\infty} v_i\).<span class="inline-marks">[3]</span></div></li></ol>

Sum to infinity

2024 VJC Promo Q4

<ol class="abc-list"><li data-list-item-id="item-1779445231871-k1145z19n" data-list-item-path="item-1779445231871-k1145z19n"><div class="list-item-paragraph">Given that&nbsp;\(\sum_{r=1}^{n} r^{2} = \frac{n}{6}(n+1)(2n+1)\), evaluate \(\sum_{r=-n}^{n} (r+1)(r+3)\) in terms of \(n\).<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1779445235135-d055hvuqe" data-list-item-path="item-1779445235135-d055hvuqe"><div class="list-item-paragraph">Using standard series from the List of Formulae (MF27), find the range of values of \(x\) for which the series \(\sum_{r=1}^{\infty} \frac{(-1)^{r+1} x^{r}}{r 2^{r}}\) converges. State the sum to infinity in terms of \(x\).<span class="inline-marks">[3]</span></div></li></ol>

2024 DHS Promo Q3

<p>It is given that&nbsp;\(\sum_{r=1}^{n} \frac{1}{r(r+1)} = 1 - \frac{1}{n+1}\).</p>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-paragraph">Find \(\sum_{r=5}^{n+1} \frac{1}{r(r+1)}\).<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b"><div class="list-item-paragraph">Give a reason why the series in part <strong>(a)</strong> is convergent and state the value of \(\sum_{r=5}^{\infty} \frac{1}{r(r+1)}\).<span class="inline-marks">[2]</span></div></li></ol>

It is given that \(\sum_{r=1}^{n} \frac{1}{r(r+1)} = 1 - \frac{1}{n+1}\).

2024 CJC Promo Q3

<ol class="i-list"><li><div class="list-item-paragraph">Find \(\sum_{r=0}^{k}\left(2r + 5 - 3^{1-r}\right)\) in terms of \(k\), simplifying your answer.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">Using the result in part <strong>(i)</strong>, find \(\sum_{r=0}^{k}\left(2r + 9 - 3^{-r-1}\right)\) in terms of \(k\).<span class="inline-marks">[3]</span></div></li></ol>

<p>A sequence \(a_1, a_2, a_3, \dots\) is such that&nbsp;\(a_n &gt; 0\), for&nbsp;\(n \in \mathbb{Z}^+\), \(n \geq 1\).</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Find \(\sum_{n=1}^{N} \ln \left( \frac{a_{n+1}}{a_n} \right)\), and express your answer as a single logarithm.<span class="inline-marks">[2]</span></div></li></ol>
<p>It is also known that this sequence is defined by \(\frac{a_{n+1}}{a_n} = \frac{1}{4}\).</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Given that&nbsp;\(\sum_{n=1}^{N} (a_n - a_{n+1})^2 = \frac{k}{16} \sum_{n=1}^{N} (a_n)^2\), find the value of \(k\).<span class="inline-marks">[2]</span></div><ol class="i-list"><li data-list-item-id="item-b-i" data-list-item-path="item-b/item-b-i"><div class="list-item-paragraph">show that&nbsp;\(\sum_{n=1}^{N} (a_n - a_{n+1})^2 = \frac{3}{5} a_1^2 \left( 1 - \frac{1}{16^N} \right)\), for&nbsp;\(N \in \mathbb{Z}^+\), \(N \geq 1\),<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b-ii" data-list-item-path="item-b/item-b-ii"><div class="list-item-paragraph">find \(\sum_{n=2}^{\infty} (a_n - a_{n+1})^2\), leaving your answers in terms of \(a_1\).<span class="inline-marks">[3]</span></div></li></ol></li></ol>

A sequence \(a_1, a_2, a_3, \dots\) is such that \(a_n > 0\), for \(n \in \mathbb{Z}^+\), \(n \geq 1\).

Find \(\sum_{n=1}^{N} \ln \left( \frac{a_{n+1}}{a_n} \right)\), and express your answer as a single logarithm.

It is also known that this sequence is defined by \(\frac{a_{n+1}}{a_n} = \frac{1}{4}\).

Given that \(\sum_{n=1}^{N} (a_n - a_{n+1})^2 = \frac{k}{16} \sum_{n=1}^{N} (a_n)^2\), find the value of \(k\).

Method of Difference

2023 ACJC Promo Q10

<p>The sum, \(S_n\), of the first \(n\) terms of the sequence \(\{u_n\}\) is given by</p>
<p style="text-align: center">\[S_n = 2n^2 - 3n, n = 1, 2, 3, \dots\]</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find the sum of the terms from the \(10\)th term to the \(23\)rd term of \(\{u_n\}\).</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find \(u_n\) in terms of \(n\).</span></div></li></ol>

The sum, \(S_n\), of the first \(n\) terms of the sequence \(\{u_n\}\) is given by

Find the sum of the terms from the \(10\)th term to the \(23\)rd term of \(\{u_n\}\).

Recovering term from sequence example

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">A sequence is such that&nbsp;\(u_1 = a\) and \(u_{n+1} = bu_n\), for&nbsp;\(n \geq 1\). Both&nbsp;\(a\) and \(b\) are constants.</div><ol class="i-list"><li data-list-item-id="item-a-i" data-list-item-path="item-a/item-a-i"><div class="list-item-paragraph">Explain why the sequence is a geometric progression.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-a-ii" data-list-item-path="item-a/item-a-ii"><div class="list-item-paragraph">Given that&nbsp;\(u_2 = 6\), \(u_1 + u_2 + u_3 = 26\), and the sum to infinity exists, find the values of \(a\) and \(b\).<span class="inline-marks">[4]</span></div></li></ol></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph"><p>It is given that&nbsp;\(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1).\)</p></div><ol class="i-list"><li data-list-item-id="item-b-i" data-list-item-path="item-b/item-b-i"><div class="list-item-paragraph">Find \(\sum_{r=1}^{n} r(r+2)\) in terms of \(n\). Leave your answer in factorised form.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-ii" data-list-item-path="item-b/item-b-ii"><div class="list-item-paragraph">Hence find \(\sum_{r=2}^{n} (r+1)(r+3)\) in terms of \(n\). You do not need to leave your answer in factorised form.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

A sequence is such that \(u_1 = a\) and \(u_{n+1} = bu_n\), for \(n \geq 1\). Both \(a\) and \(b\) are constants.

Proving the progression

2025 YIJC P2 Q4

<p>It is given that&nbsp;\(\sum_{r=2}^{n} \frac{1}{r(r^2-1)} = \frac{1}{4} + \frac{1}{2(n+1)} - \frac{1}{2n}\).</p>
<ol class="abc-list"><li data-list-item-id="item-a-4" data-list-item-path="item-a-4"><div class="list-item-paragraph">Show that&nbsp;\(\sum_{r=2}^{n} \frac{1}{r(r^2-1)}\) is less than \(\frac{1}{4}\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-4" data-list-item-path="item-b-4"><div class="list-item-paragraph">Give a reason why the series \(\sum_{r=2}^{\infty} \frac{1}{r(r^2-1)}\) converges, and write down its value.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c-4" data-list-item-path="item-c-4"><div class="list-item-paragraph">Find the smallest value of \(n\) for which \(\sum_{r=2}^{n} \frac{1}{r(r^2-1)}\) differs from \(\sum_{r=2}^{\infty} \frac{1}{r(r^2-1)}\) by less than \(0.0007\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-d-4" data-list-item-path="item-d-4"><div class="list-item-paragraph">Find \(\sum_{r=m+1}^{N} \frac{1}{(r-m)(r-m+1)(r-m+2)}\), where&nbsp;\(m\) and \(N\) are integers with \(N &gt; m &gt; 0\). (There is no need to express your answer as a single algebraic fraction.)<span class="inline-marks">[2]</span></div></li></ol>

It is given that \(\sum_{r=2}^{n} \frac{1}{r(r^2-1)} = \frac{1}{4} + \frac{1}{2(n+1)} - \frac{1}{2n}\).

Show that \(\sum_{r=2}^{n} \frac{1}{r(r^2-1)}\) is less than \(\frac{1}{4}\).

Give a reason why the series \(\sum_{r=2}^{\infty} \frac{1}{r(r^2-1)}\) converges, and write down its value.

Find the smallest value of \(n\) for which \(\sum_{r=2}^{n} \frac{1}{r(r^2-1)}\) differs from \(\sum_{r=2}^{\infty} \frac{1}{r(r^2-1)}\) by less than \(0.0007\).

Find \(\sum_{r=m+1}^{N} \frac{1}{(r-m)(r-m+1)(r-m+2)}\), where \(m\) and \(N\) are integers with \(N > m > 0\). (There is no need to express your answer as a single algebraic fraction.)

2025 VJC P1 Q6

<ol class="abc-list"><li data-list-item-id="item-a-3"><div class="list-item-content"><span class="list-item-text">Using the result \(\sum_{r=2}^{n} \frac{2}{(2r + 1)(2r + 3)} = \frac{1}{5} - \frac{1}{2n + 3}\), find \(\sum_{r=2}^{2n} \frac{2}{(2r - 1)(2r + 1)}\) in terms of \(n\).</span><span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b-3"><div class="list-item-paragraph">A sequence of positive numbers \(x_1, x_2, x_3, \dots\) satisfies the recurrence relation\[x_{n+1} = \sqrt{4 + 5x_n} \quad \text{for}\hspace{0.5em} n = 1, 2, 3, \dots\]</div><ol class="i-list"><li data-list-item-id="item-b-i-3"><div class="list-item-content"><span class="list-item-text">Given that the sequence converges to \(L\), find the exact value of \(L\).</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-ii-3"><div class="list-item-content"><span class="list-item-text">Prove that&nbsp;\(x_{n+1}^2 - L^2 = 5(x_n - L)\).</span><span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-b-iii-3"><div class="list-item-paragraph">Use the result in part <strong>(ii)</strong> to show that if&nbsp;\(x_n &gt; L\), then&nbsp;\(x_{n+1} &gt; L\).<span class="inline-marks">[2]</span></div></li></ol></li></ol>

Using the result \(\sum_{r=2}^{n} \frac{2}{(2r + 1)(2r + 3)} = \frac{1}{5} - \frac{1}{2n + 3}\), find \(\sum_{r=2}^{2n} \frac{2}{(2r - 1)(2r + 1)}\) in terms of \(n\).

2025 TJC P2 Q2

<p>It is given that&nbsp;\(\sum_{r=1}^{n} \frac{1}{r(r+1)} = 1 - \frac{1}{n+1}\).</p>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-content"><span class="list-item-text">Find \(\sum_{r=1}^{N} \frac{1}{(r+1)(r+2)}\) in terms of \(N\).</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b"><div class="list-item-paragraph">It is given that \[8 \sum_{r=k+1}^{\infty} \frac{1}{r(r+1)} = \sum_{r=1}^{k} \frac{1}{r(r+1)}.\] Find the value of \(k\).<span class="inline-marks">[3]</span></div></li></ol>

Find \(\sum_{r=1}^{N} \frac{1}{(r+1)(r+2)}\) in terms of \(N\).

It is given that\n\[8 \sum_{r=k+1}^{\infty} \frac{1}{r(r+1)} = \sum_{r=1}^{k} \frac{1}{r(r+1)}.\]\nFind the value of \(k\).

2025 RVHS P1 Q4

<p>The \(n\) th term of a series \(G\) is given by \(g_n = 3n^2 - 17n + 17\), where&nbsp;\(n \geq 1\).</p>
<ol class="abc-list"><li data-list-item-id="item-a-2" data-list-item-path="item-a-2"><div class="list-item-paragraph">Given that&nbsp;\(\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)\), find \(\sum_{r=m+1}^{n} g_r\). (You need not simplify your answer.)<span class="inline-marks">[4]</span></div></li></ol>
<p>The \(n\)th term of a series \(H\) is given by \(h_n = 7(5^{n-3}) + 10\), where&nbsp;\(n \geq 1\).</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-b-2" data-list-item-path="item-b-2"><div class="list-item-paragraph">State the smallest number that can be found in both series \(H\) and \(G\).<span class="inline-marks">[1]</span></div></li></ol>
<p>The sum of the first \(n\) terms of a series \(J\) is given by \(\frac{b}{2}(3^n - 1)\), where&nbsp;\(b\) is a positive odd integer.</p>
<ol class="abc-list" start="3"><li data-list-item-id="item-c-2" data-list-item-path="item-c-2"><div class="list-item-paragraph">By finding the \(n\)th term of series \(J\), explain why each of the terms in series \(J\) is a term of the arithmetic progression with first term \(1\) and common difference \(2\).<span class="inline-marks">[3]</span></div></li></ol>

The \(n\) th term of a series \(G\) is given by \(g_n = 3n^2 - 17n + 17\), where \(n \geq 1\).

Given that \(\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)\), find \(\sum_{r=m+1}^{n} g_r\). (You need not simplify your answer.)

The \(n\)th term of a series \(H\) is given by \(h_n = 7(5^{n-3}) + 10\), where \(n \geq 1\).

State the smallest number that can be found in both series \(H\) and \(G\).

The sum of the first \(n\) terms of a series \(J\) is given by \(\frac{b}{2}(3^n - 1)\), where \(b\) is a positive odd integer.

By finding the \(n\)th term of series \(J\), explain why each of the terms in series \(J\) is a term of the arithmetic progression with first term \(1\) and common difference \(2\).

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