A Math

A Math (IGCSE)

E Math

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

IB AI HL Math

IB AI SL Math

IP Math

PSLE

Given that $\displaystyle\sum_{r=1}^{n}\dfrac{1}{r(r+1)}=\dfrac{n}{n+1}$.

Show that $\displaystyle\sum_{r=2}^{n}\dfrac{1}{r(r-1)}=1-\dfrac{1}{n}$.

Find $\displaystyle\sum_{r=2}^{\infty}\left(\dfrac{1}{2^r}+\dfrac{1}{r(r-1)}\right)$.

Manipulating Summation Limits

Application of Sum to Infinity

Sigma Notation

2025 ACJC Promo Q3

Sequence and Series

<p>Express $\frac{r}{(2r-1)(2r+1)(2r+3)}$ in partial fractions. Hence find the sum of the first terms of the series<br></p>
<p>$S_n=\frac{1}{1\cdot 3\cdot 5}+\frac{2}{3\cdot 5\cdot 7}+\frac{3}{5\cdot 7\cdot 9}+\dots$<br>and show that&nbsp;$S_n&lt;\frac{1}{8}$.</p>

Express $\frac{r}{(2r-1)(2r+1)(2r+3)}$ in partial fractions. Hence find the sum of the first terms of the series


$S_n=\frac{1}{1\cdot 3\cdot 5}+\frac{2}{3\cdot 5\cdot 7}+\frac{3}{5\cdot 7\cdot 9}+\dots$
and show that $S_n<\frac{1}{8}$.

Application of Sum to Infinity Q2

<ol class="i-list"><li><div class="list-item-paragraph">Find $\sum\limits_{r=1}^{n}{\frac{1}{r(r+2)}}$ in terms of $n$.</div></li><li><div class="list-item-paragraph">Give a reason why $\sum\limits_{r=1}^{\infty }{\frac{1}{r(r+2)}}$ converges and write down its value.</div></li></ol>

Application of Sum to Infinity Q1

<p>Amil and Babs are playing a game in which they take turns to throw a fair six-sided die. The winner is the first player to throw the same number as the other player has thrown immediately before. Anil throws first.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Find exactly the probability that Anil wins the game.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Given that Babs wins the game, find exactly the probability that she wins on her second throw.</div><p style="text-align: right;">[3]</p></li></ol>

Amil and Babs are playing a game in which they take turns to throw a fair six-sided die. The winner is the first player to throw the same number as the other player has thrown immediately before. Anil throws first.

Conditional Probability

Probability of a Sequence

N2022 P2 Q6

Probability

<p>By considering the series expansions of ${\ln \left( 1-{{x}^{2}} \right)}$ and ${\ln \left( \frac{1+x}{1-x} \right)}$, or otherwise, find the exact value of the following series.</p>
<p style="text-align: center">${\sum\limits_{r=1}^{\infty }{\left[ \left( \frac{1}{2r}+\frac{1}{2r+1} \right){{\left( \frac{1}{4} \right)}^{r}} \right]}}$.</p>

By considering the series expansions of ${\ln \left( 1-{{x}^{2}} \right)}$ and ${\ln \left( \frac{1+x}{1-x} \right)}$, or otherwise, find the exact value of the following series.

${\sum\limits_{r=1}^{\infty }{\left[ \left( \frac{1}{2r}+\frac{1}{2r+1} \right){{\left( \frac{1}{4} \right)}^{r}} \right]}}$.

Maclaurin Series Using Standard Series

Maclaurin and Power Series Q1

Maclaurin Series and Power Series

<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Prove by method of differences that $\sum\limits_{r=1}^{n}{\frac{1}{\left( 2r-1 \right)\left( 2r+3 \right)}}=\frac{1}{3}-\frac{n+1}{\left( 2n+1 \right)\left( 2n+3 \right)}$ .</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">Hence, find</span></div><ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">the exact value of $\sum\limits_{r=5}^{\infty }{\frac{1}{\left( 2r-1 \right)\left( 2r+3 \right)}}$,</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">$\sum\limits_{r=0}^{n}{\frac{1}{\left( 2r+1 \right)\left( 2r+5 \right)}}$in terms of $n$.</span></div><p style="text-align: right;">[2]</p></li></ol></li></ol>

Prove by method of differences that $\sum\limits_{r=1}^{n}{\frac{1}{\left( 2r-1 \right)\left( 2r+3 \right)}}=\frac{1}{3}-\frac{n+1}{\left( 2n+1 \right)\left( 2n+3 \right)}$ .

Method of Difference

2015 NJC P1 Q5

<ol class="i-list"><li><div class="list-item-paragraph">By considering $\mathrm{f}\left( r \right)-\mathrm{f}\left( r+1 \right)$, where $\mathrm{f}\left( r \right)=\frac{1}{r}$, find the sum to $2n$ terms of the series $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...$</div></li><li><div class="list-item-paragraph">Show that the sum to $2n$ terms of the series $\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}+...$ is less than $1$.</div></li></ol>

By considering $\mathrm{f}\left( r \right)-\mathrm{f}\left( r+1 \right)$, where $\mathrm{f}\left( r \right)=\frac{1}{r}$, find the sum to $2n$ terms of the series $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...$

Show that the sum to $2n$ terms of the series $\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}+...$ is less than $1$.

Application of Sum to Infinity Q3

<p>Write down the first three terms in ascending powers of $x$ of the expansion of ${{\left( 1-x \right)}^{-2}}$. State the coefficient of ${{x}^{n}}$ and the range of values of $x$ for which the expansion is valid. Hence, or otherwise, find the value of $\sum_{n=1}^{\infty} \frac{n}{2^n}$.<span class="inline-marks">[5]</span></p>
<p style="text-align: center"></p>

Write down the first three terms in ascending powers of $x$ of the expansion of ${{\left( 1-x \right)}^{-2}}$. State the coefficient of ${{x}^{n}}$ and the range of values of $x$ for which the expansion is valid. Hence, or otherwise, find the value of $\sum_{n=1}^{\infty} \frac{n}{2^n}$.

Math Question Bank