A Math

A Math (IGCSE)

E Math

E Math (IGCSE)

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

IB AI HL Math

IB AI SL Math

IP Math

N(A) A Math

N(A) E Math

PSLE

PSLE Foundation Mathematics

<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 $n$.

Manipulating Summation Limits

Summing Polynomial Series

2025 YIJC Promo Q4

Sequence and Series

<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 $.

Sigma Notation By Substitution

Problems involving inequalities

Geometric Progression

2025 VJC Promo Q3

<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>

Limit of a Sequence and Behaviour of a Sequence

Generating Terms from Recurrence Relations

2025 ACJC P2 Q3

<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-a"><div class="list-item-paragraph">Given that&nbsp;$\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)$, find $\sum_{r=1}^{n} \left[ \left(\frac{1}{2}\right)^r - r\left(r+\frac{1}{3}\right) \right]$, in terms of $n$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b"><div class="list-item-paragraph">Explain why $\sum_{r=1}^{n} \left[ \left(\frac{1}{2}\right)^r - r\left(r+\frac{1}{3}\right) \right] &lt; 1$ for all positive integers $n$.<span class="inline-marks">[1]</span></div></li></ol>

2024 HCI Promo Q2

<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

<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

Sum to infinity

2025 YIJC P2 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 $2$.

Inter-sequence Relationships

$u_n = S_n - S_{n-1} $

2025 NYJC P1 Q5

<p>A series is given by $\sum_{r=1}^{n} 2(4-3x)^r$ where&nbsp;$x$ is constant.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Explain why this is a geometric series. Determine the range of values of $x$ for the sum to infinity of this series to exist.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Using $x=1$, and given that&nbsp;$\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)$, find $\sum_{r=0}^{n-1} [2(4-3x)^r + (r+1)(2r+5)]$, leaving your answer in the form of $n(an^2+bn+c)$, where&nbsp;$a$, $b$ and $c$ are constants to be determined.<span class="inline-marks">[4]</span></div></li></ol>

A series is given by $\sum_{r=1}^{n} 2(4-3x)^r$ where $x$ is constant.

Explain why this is a geometric series. Determine the range of values of $x$ for the sum to infinity of this series to exist.

Using $x=1$, and given that $\sum_{r=1}^{n} r^2 = \frac{n}{6}(n+1)(2n+1)$, find $\sum_{r=0}^{n-1} [2(4-3x)^r + (r+1)(2r+5)]$, leaving your answer in the form of $n(an^2+bn+c)$, where $a$, $b$ and $c$ are constants to be determined.

2025 DHS P1 Q2 

<p>By considering the expansion of ${(r+1)}^{3}$, show that&nbsp;$\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)$.</p>

By considering the expansion of ${(r+1)}^{3}$, show that $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)$.

Method of Difference

Sigma Notation Practice 1 Q4

<p>Given that $\sum\limits_{r=1}^{n}{r}=\frac{n}{2}(n+1)$, $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)$ and $\sum\limits_{r=1}^{n}{{{r}^{3}}}=\frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}={{\left[ \frac{1}{2}n\left( n+1 \right) \right]}^{2}}$.</p>
<p>Find, in terms of $n$,</p>
<ol class="i-list"><li><div class="list-item-paragraph">$\sum\limits_{r=0}^{n}{\left( {{r}^{2}}+r-n \right)}$,</div></li><li><div class="list-item-paragraph">$\sum\limits_{r=1}^{20}{\left( 2{{r}^{2}}-1 \right)\left( r+1 \right)}$.</div></li></ol>

Given that $\sum\limits_{r=1}^{n}{r}=\frac{n}{2}(n+1)$, $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)$ and $\sum\limits_{r=1}^{n}{{{r}^{3}}}=\frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}={{\left[ \frac{1}{2}n\left( n+1 \right) \right]}^{2}}$.

Change of Base Q3

<p>Given that $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. Find, in terms of $n$,</p>
<ol class="i-list"><li><div class="list-item-paragraph">$\sum\limits_{r=-2}^{n}{{{r}^{2}}}$</div></li><li><div class="list-item-paragraph">$\sum\limits_{r=-3}^{n+1}{{{r}^{2}}}$</div></li><li><div class="list-item-paragraph">$\sum\limits_{r=-2}^{n-3}{2{{r}^{2}}}$</div></li></ol>

Given that $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. Find, in terms of $n$,

Change of Base Q2

<ol class="abc-list"><li><div class="list-item-paragraph">Find $\sum\limits_{r=0}^{n}{\left[ \left( n+2 \right)r+n \right]}$, giving your answer in terms of $n$.</div><p style="text-align: right;">[3]</p></li></ol>
<p>[You may use the result $\sum\limits_{r=1}^{n}{{{r}^{3}}=\frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}}$ for the rest of this question.]</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">By writing down the first two and the last two terms in the series, find $\sum\limits_{r=1}^{n}{{{\left( r+2 \right)}^{3}}}$, giving your answer in terms of $n$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Find ${{1}^{3}}-{{2}^{3}}+{{3}^{3}}-{{4}^{3}}+{{5}^{3}}-{{6}^{3}}+\ldots +{{\left( 2n-1 \right)}^{3}}-{{\left( 2n \right)}^{3}}$, giving your answer in terms of $n$.</div><p style="text-align: right;">[3]</p></li></ol>

[You may use the result $\sum\limits_{r=1}^{n}{{{r}^{3}}=\frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}}$ for the rest of this question.]

Arithmetic Progression

Sigma Notation

2024 RI Promo Q5

<ol class="abc-list"><li><div class="list-item-paragraph">For this question, you may use these results:</div><div class="list-item-paragraph" style="text-align: center">$\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{n(n+1)(2n+1)}{6}$ and $\sum\limits_{r=1}^{n}{{{r}^{3}}}=\frac{{{n}^{2}}{{(n+1)}^{2}}}{4}$.</div><ol class="i-list"><li><div class="list-item-paragraph">Show that $\sum\limits_{r=1}^{n}{r}{{(r+1)}^{2}}=\frac{n(n+1)(n+2)(3n+5)}{12}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Hence find $\sum\limits_{r=5}^{n-1}{(r+2)}{{(r+3)}^{2}}$ in terms of $n$.</div><p style="text-align: right;">[3]</p></li></ol></li><li><div class="list-item-paragraph">The sequence ${{u}_{1}},{{u}_{2}},{{u}_{3}},\ldots $ 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 ${{u}_{2025}}$.</div><p style="text-align: right;">[3]</p></li></ol>

<p>The shaded region ${R}$ is bounded by the ${x}$-axis, the ${y}$-axis, the line $y=1$ and the curve ${x={{y}^{2}}+1}$. There are ${n}$ rectangles of equal width drawn as shown in the diagram below. The total area of all the n rectangles, ${{A}_{n}}$ approximates the area of the shaded region ${R}$.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760850618625-0dlyl06u5nqq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA4NTA2MTg2MjUtMGRseWwwNnU1bnFxLnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTI3NjM1NjAsImV4cCI6MTc5Mjc2Mzg2MH0.65DmvweMcgM8pEANVatQs0W5xr4FtLE6_RgxwETlUFk" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>It is given that&nbsp;${\mathop{\sum }_{r=1}^{n}{{r}^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}$.</p>
<ol class="abc-list"><li data-list-item-id="item-1760850620272-6wovwoy1f" data-list-item-path="item-1760850620272-6wovwoy1f"><div class="list-item-paragraph">Show that&nbsp;${{{A}_{n}}=1+\frac{\left( n-1 \right)\left( 2n-1 \right)}{6{{n}^{2}}}}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1760850620895-hx06injyr" data-list-item-path="item-1760850620895-hx06injyr"><div class="list-item-paragraph">Find the exact value of ${\mathop{\lim}\limits_{n\to \infty }\,{{A}_{n}}}$, showing your working.<span class="inline-marks">[3]</span></div></li></ol>

The shaded region ${R}$ is bounded by the ${x}$-axis, the ${y}$-axis, the line $y=1$ and the curve ${x={{y}^{2}}+1}$. There are ${n}$ rectangles of equal width drawn as shown in the diagram below. The total area of all the n rectangles, ${{A}_{n}}$ approximates the area of the shaded region ${R}$.

It is given that ${\mathop{\sum }_{r=1}^{n}{{r}^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}$.

Limit of an expression

Riemann Sums

2023 HCI Promo Q4

Applications of Integration

<ol class="abc-list"><li><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/converted/1760753909486-xr3u9j011k.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvY29udmVydGVkLzE3NjA3NTM5MDk0ODYteHIzdTlqMDExay5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzYzNTYwLCJleHAiOjE3OTI3NjM4NjB9.bqnVnTP_10zJ_IexbyCRAP5UT0hh2PFdb14vf8589cc" alt="List item image" class="aligncenter custom-width" style="max-width: min(400px, 100%); height: auto;" /></div><div class="list-item-paragraph">The diagram shows part of the graph of $y={{x}^{2}}+2$ from $x=1$ to $x=2$. The area under the curve in this interval may be approximated by the total area of $n$ rectangles, $A$, as shown. The width of each rectangle is $\frac{1}{n}$.</div><div class="list-item-paragraph">Given that&nbsp;$\underset{r=1}{\overset{n}{\mathop \sum }}\,{{r}^{2}}=\frac{n}{6}\left( n+1 \right)\left( 2n+1 \right)$, show that&nbsp;$A=\frac{13}{3}+\frac{3}{2n}+\frac{1}{6{{n}^{2}}}$.</div><div class="list-item-paragraph">Deduce the exact value of $\mathop{\displaystyle\displaystyle \displaystyle\int }_{1}^{2}\left( 2+{{x}^{2}} \right)\,\mathrm{d}x$, justifying your answer. <span class="inline-marks">[6]</span></div></li><li><div class="list-item-paragraph">Region $R$ is bounded by the curve $y=\ln x$, $x=1$, and $y=\ln 5$.</div><ol class="i-list"><li><div class="list-item-paragraph">Find the area of $R$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find the exact volume of the solid generated when&nbsp;$R$ is rotated through $4$ right angles about the $y$-axis.<span class="inline-marks">[4]</span></div></li></ol></li></ol>

Integration by Parts

Volume of Revolution about the y-axis

Area Enclosed by a Curve and the y-axis

2024 CJC P1 Q9

Techniques of Integration

<p>A sequence of numbers ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … has a sum ${{S}_{n}}$ where ${{S}_{n}}=\sum\limits_{r=1}^{n}{{{u}_{r}}}$. It is given that ${{S}_{n}}=A{{n}^{2}}\left( n+1 \right)-Bn$, where $A$ and $B$ are non-zero constants.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find a simplified expression for ${{u}_{n}}$ in terms of $A$, $B$ and $n$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">It is also given that the first and second term of the sequence are $1$ and $9$ respectively. Show that $A=1$ and $B=1$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Hence, find $\sum\limits_{r=1}^{n}{\left[ 3{{\left( r-1 \right)}^{2}}-r \right]}$, giving your answer in the form $a{{n}^{3}}-b{{n}^{2}}$, where $a$ and $b$ are constants to be determined.</div><p style="text-align: right;">[3]</p></li></ol>

A sequence of numbers ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … has a sum ${{S}_{n}}$ where ${{S}_{n}}=\sum\limits_{r=1}^{n}{{{u}_{r}}}$. It is given that ${{S}_{n}}=A{{n}^{2}}\left( n+1 \right)-Bn$, where $A$ and $B$ are non-zero constants.

2020 NYJC P1 Q7

<ol class="abc-list"><li data-list-item-id="item-1760681178336-rnkfyw3g4" data-list-item-path="item-1760681178336-rnkfyw3g4"><div class="list-item-paragraph">Given that&nbsp;$\sum\limits_{r=1}^{n}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)}$, find $\sum\limits_{r=1}^{n}{6r\left( r+1 \right)}$, in terms of $n$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1760681179168-1k2wy19wd" data-list-item-path="item-1760681179168-1k2wy19wd"><ol class="i-list"><li data-list-item-id="subitem-1760681189616-he5nm1l0a" data-list-item-path="item-1760681179168-1k2wy19wd/subitem-1760681189616-he5nm1l0a"><div class="list-item-paragraph">Express $\frac{1}{r\left( r+1 \right)\left( r+2 \right)}$ in partial fractions.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="subitem-1760681190718-m7jx45bh4" data-list-item-path="item-1760681179168-1k2wy19wd/subitem-1760681190718-m7jx45bh4"><div class="list-item-paragraph">Hence find ${{S}_{N}}$ in terms of $N$, where&nbsp;${{S}_{N}}=\sum\limits_{r=1}^{N}{\frac{1}{r(r+1)(r+2)}}$, and deduce that&nbsp;${{S}_{N}}&lt;\frac{1}{4}$.<span class="inline-marks">[6]</span></div></li></ol></li></ol>

2023 SAJC Promo Q3

<p>It is given that $\sum\limits_{r=\,\,1}^{n}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that $\sum\limits_{r=\,\,1}^{n}{\left( 2r-7 \right)\left( r+1 \right)}=\frac{1}{6}n\left( 4{{n}^{2}}-9n-55 \right)$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Find $\sum\limits_{r=\,\,1}^{n}{{{3}^{-\,r}}}$ in terms of $n$, and find the least value of $n$ such that $\sum\limits_{r=\,\,1}^{n}{\left( 2r-7 \right)}\left( r+1 \right)>\sum\limits_{r=\,\,1}^{n}{{{3}^{-\,r}}}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Express $\sum\limits_{r=\,\,n+1}^{2n}{\left( 2r-7 \right)\left( r+1 \right)}$ in terms of $n$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Hence, or otherwise, find the value of $43\times 26+45\times 27+47\times 28+...87\times 48+89\times 49$.</div><p style="text-align: right;">[2]</p></li></ol>

It is given that $\sum\limits_{r=\,\,1}^{n}{{{r}^{2}}=\frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right)}$.

2022 EJC Promo Q11

<ol class="abc-list"><li><ol class="i-list"><li><div class="list-item-paragraph">Using the method of difference, show that&nbsp;$\sum\limits_{r=\,\,1}^{n}{\left[ {{\left( r+1 \right)}^{3}}-{{r}^{3}} \right]}={{\left( n+1 \right)}^{3}}-1$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Expand and simplify ${{\left( r+1 \right)}^{3}}-{{r}^{3}}$.<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">Hence, by using <strong>(i)</strong> and <strong>(ii)</strong>, show that</div><div class="list-item-paragraph" style="text-align: center">$\sum\limits_{r=\,\,1}^{n}{{{r}^{2}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}$.<span class="inline-marks">[3]</span></div></li></ol></li><li><div class="list-item-paragraph">It is given that&nbsp;$\sum\limits_{r=\,\,1}^{n}{\frac{2r+1}{r\left( r+1 \right)\left( r+2 \right)}=\frac{n\left( 5n+7 \right)}{4\left( n+1 \right)\left( n+2 \right)}}$.</div><ol class="i-list"><li><div class="list-item-paragraph">Show that the series $\sum\limits_{r=\,\,1}^{\infty }{\frac{2r+1}{r\left( r+1 \right)\left( r+2 \right)}}$ converges and write down its sum to infinity.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Find $\sum\limits_{r=\,0}^{n-2}{\frac{2r+5}{\left( r+2 \right)\left( r+3 \right)\left( r+4 \right)}}$.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

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