# Sigma Notation

###### 5 Essential Questions

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• Q2
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• Q5
##### 2017 EJC Promo Q7 (b) [Modified]

Given that $\sum\limits_{r=2}^{n}{\frac{1}{{{r}^{2}}-1}}=\frac{3}{4}-\frac{1}{2n}-\frac{1}{2\left( n+1 \right)}$,

(ii)

state $\sum\limits_{r=2}^{\infty }{\frac{1}{{{r}^{2}}-1}}$,

(ii) state $\sum\limits_{r=2}^{\infty }{\frac{1}{{{r}^{2}}-1}}$,

(iii)

find $\sum\limits_{r=2}^{n-1}{\frac{1}{r(r+2)}}$.

(iii) find $\sum\limits_{r=2}^{n-1}{\frac{1}{r(r+2)}}$.

• (ii)
• (iii)

• (ii)
• (iii)

##### 2018 YJC Promo Q13

The terms of a geometric progression ${{u}_{1}},{{u}_{2}},{{u}_{3}},{{u}_{4}},…$ are such that the sum to infinity is $81$ and the sum of the first $4$ terms is $80$.
If ${{u}_{1}}>100\,\,$and $\,\text{n}\ge \text{3}$,

(i)

Show that $\frac{3}{r}-\frac{6}{r+1}+\frac{3}{r+2}=\frac{6}{r\left( r+1 \right)\left( r+2 \right)}$.

[1]

(i) Show that $\frac{3}{r}-\frac{6}{r+1}+\frac{3}{r+2}=\frac{6}{r\left( r+1 \right)\left( r+2 \right)}$.

[1]

(ii)

Hence show that $\sum\limits_{r=1}^{N}{\frac{1}{r\left( r+1 \right)\left( r+2 \right)}}=\frac{1}{4}-\frac{1}{2\left( N+1 \right)}+\frac{1}{2\left( N+2 \right)}$.

[3]

(ii) Hence show that $\sum\limits_{r=1}^{N}{\frac{1}{r\left( r+1 \right)\left( r+2 \right)}}=\frac{1}{4}-\frac{1}{2\left( N+1 \right)}+\frac{1}{2\left( N+2 \right)}$.

[3]

(iii)

Give a reason why the series $\sum\limits_{r=1}^{\infty }{\frac{1}{r\left( r+1 \right)\left( r+2 \right)}}$ converges and write down its value.

[2]

(iii) Give a reason why the series $\sum\limits_{r=1}^{\infty }{\frac{1}{r\left( r+1 \right)\left( r+2 \right)}}$ converges and write down its value.

[2]

(iv)

Use your answer to (ii) to find $\sum\limits_{r=3}^{N}{\frac{1}{r\left( r-1 \right)\left( r-2 \right)}}$in terms of $N$.

[2]

(iv) Use your answer to (ii) to find $\sum\limits_{r=3}^{N}{\frac{1}{r\left( r-1 \right)\left( r-2 \right)}}$in terms of $N$.

[2]

• (i)
• (ii)
• (iii)
• (iv)

• (i)
• (ii)
• (iii)
• (iv)

##### 2019 DHS Promo Q2

Using the result $\sum\limits_{r=1}^{n}{\frac{r}{{{2}^{r}}}}=2-\frac{n+2}{{{2}^{n}}}$, show that $\sum\limits_{r=1}^{n}{\left( r-n \right)\left( {{2}^{-r}}+1 \right)}$ can be expressed in the form $C\left( 1-\frac{1}{{{2}^{n}}} \right)+Dn\left( n+1 \right)$, where $C$ and $D$ are constants to be determined.

[4]

##### 2017 VJC P1 Q8

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

(i)

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

[3]

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

[3]

(ii)

Show that $\sum\limits_{r=4}^{n}{\frac{{{\left( r-2 \right)}^{2}}}{{{3}^{r-2}}}}=\frac{p}{q}-\frac{a{{n}^{2}}-an+a}{2\left( {{3}^{n-2}} \right)}$, where $a$, $p$ and $q$ are integers to be determined.

[5]

(ii) Show that $\sum\limits_{r=4}^{n}{\frac{{{\left( r-2 \right)}^{2}}}{{{3}^{r-2}}}}=\frac{p}{q}-\frac{a{{n}^{2}}-an+a}{2\left( {{3}^{n-2}} \right)}$, where $a$, $p$ and $q$ are integers to be determined.

[5]

• (i)
• (ii)
• (i)
• (ii)

##### 2020 EJC P1 Q10

The diagram shows the graph of $y=\frac{1}{{{x}^{2}}+1}$ when $x>0$.

(i)

Evaluate $\int_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\text{d}x}$ for $k>0$, leaving your answer in terms of $k$.

[2]

(i) Evaluate $\int_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\text{d}x}$ for $k>0$, leaving your answer in terms of $k$.

[2]

(ii)

By considering appropriate rectangles on the interval $\left[ k,k+1 \right]$for the curve $y=\frac{1}{{{x}^{2}}+1}$, show that

$\frac{1}{{{\left( k+1 \right)}^{2}}+1}<{{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}k<\frac{1}{{{k}^{2}}+1}$ for $k\in {{\mathbb{Z}}^{+}}$.

[2]

(ii) By considering appropriate rectangles on the interval $\left[ k,k+1 \right]$for the curve $y=\frac{1}{{{x}^{2}}+1}$, show that

$\frac{1}{{{\left( k+1 \right)}^{2}}+1}<{{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}k<\frac{1}{{{k}^{2}}+1}$ for $k\in {{\mathbb{Z}}^{+}}$.

[2]

(iii)

Use the identity $\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ to show that

${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\frac{x-y}{1+xy}$, where $x>y>0$.

[2]

(iii) Use the identity $\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ to show that

${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\frac{x-y}{1+xy}$, where $x>y>0$.

[2]

(iv)

By considering parts (ii) and (iii), prove by the method of differences that

$\sum\limits_{k=1}^{n}{\frac{1}{{{\left( k+1 \right)}^{2}}+1}}<{{\tan }^{-1}}\left( \frac{n}{n+2} \right)<\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+1}}$

[4]

(iv) By considering parts (ii) and (iii), prove by the method of differences that

$\sum\limits_{k=1}^{n}{\frac{1}{{{\left( k+1 \right)}^{2}}+1}}<{{\tan }^{-1}}\left( \frac{n}{n+2} \right)<\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+1}}$

[4]

• (i)
• (ii)
• (iii)
• (iv)
• (i)
• (ii)
• (iii)
• (iv)