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<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.

Limit of a Sequence and Behaviour of a Sequence

Method of Difference

Sigma Notation

2026 ASRJC P1 Q10

Sequence and Series

<p>Express $\frac{{{r}^{2}}+7r+11}{\left( r+4 \right)!}$ in the form $\frac{A}{\left( r+2 \right)!}+\frac{B}{\left( r+4 \right)!}$ where&nbsp;$A$ and $B$ are constants to be found.</p>
<p>Hence show that&nbsp;$\sum\limits_{r=1}^{n}{\frac{{{r}^{2}}+7r+11}{\left( r+4 \right)!}}=\frac{5}{24}-\frac{n+5}{\left( n+4 \right)!}$.</p>

Express $\frac{{{r}^{2}}+7r+11}{\left( r+4 \right)!}$ in the form $\frac{A}{\left( r+2 \right)!}+\frac{B}{\left( r+4 \right)!}$ where $A$ and $B$ are constants to be found.

Hence show that $\sum\limits_{r=1}^{n}{\frac{{{r}^{2}}+7r+11}{\left( r+4 \right)!}}=\frac{5}{24}-\frac{n+5}{\left( n+4 \right)!}$.

Method of Difference Q5

<p>Use the method of differences to find $\sum\limits_{r=1}^{n}{\ln \left( \frac{r}{r+2} \right)}$ in the form $\ln \left[ \frac{k}{\left( n+1 \right)\left( n+2 \right)} \right]$, where&nbsp;$k$ is a constant to be determined.</p>
<p>Hence find $\ln \left( \frac{n+1}{n+3} \right)+\ln \left( \frac{n+2}{n+4} \right)+...+\ln \left( \frac{3n}{3n+2} \right)$.</p>

Use the method of differences to find $\sum\limits_{r=1}^{n}{\ln \left( \frac{r}{r+2} \right)}$ in the form $\ln \left[ \frac{k}{\left( n+1 \right)\left( n+2 \right)} \right]$, where $k$ is a constant to be determined.

Hence find $\ln \left( \frac{n+1}{n+3} \right)+\ln \left( \frac{n+2}{n+4} \right)+...+\ln \left( \frac{3n}{3n+2} \right)$.

Method of Difference Q4

<p>Find the sum of the series $\sum\limits_{r=2}^{n}{\ln \left( \frac{r}{r-1} \right)}$.</p>

Find the sum of the series $\sum\limits_{r=2}^{n}{\ln \left( \frac{r}{r-1} \right)}$.

Method of Difference Q3

<p>By expressing $\frac{1}{(3r-2)(3r+1)}$ in partial fractions, evaluate $\sum\limits_{r=1}^{n}{\frac{1}{(3r-2)(3r+1)}}$.</p>

By expressing $\frac{1}{(3r-2)(3r+1)}$ in partial fractions, evaluate $\sum\limits_{r=1}^{n}{\frac{1}{(3r-2)(3r+1)}}$.

Method of Difference Q2

<p>By expressing $\frac{1}{(4r-1)(4r+3)}$ in partial fractions, evaluate $\sum\limits_{r=1}^{n}{\frac{1}{(4r-1)(4r+3)}}$.</p>

By expressing $\frac{1}{(4r-1)(4r+3)}$ in partial fractions, evaluate $\sum\limits_{r=1}^{n}{\frac{1}{(4r-1)(4r+3)}}$.

Method of Difference Q1

<p>Express $\frac{1}{(x+1)(x+2)}$ in partial fractions.</p>
<p>Hence,</p>
<ol class="i-list"><li><div class="list-item-paragraph">Evaluate $\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+........+\frac{1}{(19)(20)}$.</div></li><li><div class="list-item-paragraph">Find the value of $n$ if $\frac{1}{(1)(2)}+\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+........+\frac{1}{(n-1)(n)}=\frac{99}{100}$.</div></li></ol>

Express $\frac{1}{(x+1)(x+2)}$ in partial fractions.

Partial Fractions

Question 114081

Quadratic Functions, Equations and Inequalities

Polynomials, Cubic Equations and Partial Fractions

<p>Let ${{u}_{n}}=\ln \left( \frac{n}{n+1} \right)+\frac{1}{n+1}-\frac{1}{n}$ ,</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find in terms of $N$, an expression for ${{S}_{N}}$, where ${{S}_{N}}={{u}_{1}}+{{u}_{2}}+...+{{u}_{N}}$, simplifying your result as far as possible.</div></li><li><div class="list-item-paragraph">Show that ${{S}_{N}}<0$ for all $N\ge 1$.</div></li></ol>

Let ${{u}_{n}}=\ln \left( \frac{n}{n+1} \right)+\frac{1}{n+1}-\frac{1}{n}$ ,

AJC P1 Q8 (a)

<p>A sequence ${{u}_{0}}$, ${{u}_{1}}$, ${{u}_{2}}$… is such that&nbsp;${{u}_{r}}=\frac{1}{r!}$ and ${{u}_{r}}={{u}_{r-2}}+\frac{r+1-{{r}^{2}}}{r!}$, when&nbsp;$n\ge 2$.</p>
<ol class="i-list"><li data-list-item-id="item-1759737554530-cznhathnh" data-list-item-path="item-1759737554530-cznhathnh"><div class="list-item-paragraph">Show that&nbsp;$\sum\limits_{r=2}^{n}{\frac{{{r}^{2}}-r-1}{r!}}=2-\frac{n+1}{n!}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1759737555385-gdricbxr9" data-list-item-path="item-1759737555385-gdricbxr9"><div class="list-item-paragraph">Hence find $\sum\limits_{r=8}^{n+5}{\frac{{{r}^{2}}-3r+1}{\left( r-1 \right)!}}$ in terms of $n$.<span class="inline-marks">[3]</span></div></li></ol>
<p>Limit Comparison test states that for two series of the form $\sum\limits_{r=k}^{n}{{{a}_{r}}}$ and $\sum\limits_{r=k}^{n}{{{b}_{r}}}$ with ${{a}_{n}},{{b}_{n}}\ge 0$ for all $n$, if&nbsp;$\mathop{\lim}\limits_{n\to \infty }\,\frac{{{a}_{n}}}{{{b}_{n}}}&gt;0$, then both&nbsp;$\sum\limits_{r=k}^{\infty }{{{a}_{r}}}$and $\sum\limits_{r=k}^{\infty }{{{b}_{r}}}$ converges or both diverges.</p>
<ol class="i-list" start="3"><li data-list-item-id="item-1759737581942-t71zjddfx" data-list-item-path="item-1759737581942-t71zjddfx"><div class="list-item-paragraph">Given that&nbsp;$\sum\limits_{r=2}^{\infty }{\frac{{{r}^{2}}-r-1}{r!}}$ is convergent, using the test, explain why $\sum\limits_{r=2}^{\infty }{\frac{r-2}{\left( r-1 \right)!}}$ is convergent.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1759737582662-woqdbtvjv" data-list-item-path="item-1759737582662-woqdbtvjv"><div class="list-item-paragraph">Show that&nbsp;$e-2&lt;\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...&lt;2$.<span class="inline-marks">[3]</span></div></li></ol>

A sequence ${{u}_{0}}$, ${{u}_{1}}$, ${{u}_{2}}$… is such that ${{u}_{r}}=\frac{1}{r!}$ and ${{u}_{r}}={{u}_{r-2}}+\frac{r+1-{{r}^{2}}}{r!}$, when $n\ge 2$.

Limit Comparison test states that for two series of the form $\sum\limits_{r=k}^{n}{{{a}_{r}}}$ and $\sum\limits_{r=k}^{n}{{{b}_{r}}}$ with ${{a}_{n}},{{b}_{n}}\ge 0$ for all $n$, if $\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{a}_{n}}}{{{b}_{n}}}>0$, then both $\sum\limits_{r=k}^{\infty }{{{a}_{r}}}$and $\sum\limits_{r=k}^{\infty }{{{b}_{r}}}$ converges or both diverges.

Manipulating Summation Limits

2018 SRJC P1 Q8

<ol class="i-list"><li><div class="list-item-paragraph">Show that $\frac{4}{\left( r-1 \right)!}-\frac{1}{r!}-\frac{3}{\left( r+1 \right)!}=\frac{4{{r}^{2}}+3r-4}{\left( r+1 \right)!}$ .</div><div class="list-item-paragraph">Hence, express $\sum\limits_{r=1}^{n}{\frac{4{{r}^{2}}+3r-4}{\left( r+1 \right)!}}$ in the form $A-\mathrm{f}\left( n \right)$, where $A$ is a constant and $\mathrm{f}\left( n \right)$ is a single fraction in terms of $n$.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-paragraph">Deduce that $\sum\limits_{r=1}^{n}{\frac{{{r}^{2}}-1}{\left( r+1 \right)!}<\frac{7}{4}}$ .</div><p style="text-align: right;">[3]</p></li></ol>

2018 EJC PROMO Q8

<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 $\sum\limits_{r=1}^{n}{\frac{1}{\left( 2r-1 \right)\left( 2r+3 \right)}}=\frac{1}{3}-\frac{n+1}{\left( 2n+1 \right)\left( 2n+3 \right)}$ .

Application of Sum to Infinity

2015 NJC P1 Q5

<p>Given that $\mathrm{f}\left( r \right)=\frac{r}{{{2}^{r}}}$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">By considering $\mathrm{f}\left( r+1 \right)-\mathrm{f}\left( r \right)$, find $\sum\limits_{r=1}^{n}{\frac{1-r}{{{2}^{r+1}}}}$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Find $\sum\limits_{r=3}^{n}{\frac{r}{{{2}^{r+2}}}}$.<span class="inline-marks">[3]</span></div></li></ol>

Given that $\mathrm{f}\left( r \right)=\frac{r}{{{2}^{r}}}$.

By considering $\mathrm{f}\left( r+1 \right)-\mathrm{f}\left( r \right)$, find $\sum\limits_{r=1}^{n}{\frac{1-r}{{{2}^{r+1}}}}$.

Find $\sum\limits_{r=3}^{n}{\frac{r}{{{2}^{r+2}}}}$.

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