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<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 $\{u_n\}$.

Method of Difference involving uₙ−uₙ₋₁.

Recovering term from sequence example

Sequence and Series

<ol class="i-list"><li data-list-item-id="item-1759486649461-rwv4qa964" data-list-item-path="item-1759486649461-rwv4qa964"><div class="list-item-paragraph">By writing $\frac{2-r}{r\left( r+1 \right)\left( r+2 \right)}$ in partial fractions, show that&nbsp;$\sum\limits_{r=1}^{n}{\frac{2-r}{r\left( r+1 \right)\left( r+2 \right)}=\frac{An}{\left( n+1 \right)\left( n+2 \right)}}$, where&nbsp;$A$ is a constant to be determined.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1759486649979-ffwj3htmz" data-list-item-path="item-1759486649979-ffwj3htmz"><div class="list-item-paragraph">Using the result in part <strong>(i)</strong>, find $\sum\limits_{r=0}^{n}{\frac{1-r}{\left( r+1 \right)\left( r+2 \right)\left( r+3 \right)}}$ in terms of $n$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1759486650676-8moz9nan7" data-list-item-path="item-1759486650676-8moz9nan7"><div class="list-item-paragraph">Hence find the exact value of $\sum\limits_{r=10}^{\infty }{\frac{1-r}{\left( r+1 \right)\left( r+2 \right)\left( r+3 \right)}}$.<span class="inline-marks">[2]</span></div></li></ol>

By writing $\frac{2-r}{r\left( r+1 \right)\left( r+2 \right)}$ in partial fractions, show that $\sum\limits_{r=1}^{n}{\frac{2-r}{r\left( r+1 \right)\left( r+2 \right)}=\frac{An}{\left( n+1 \right)\left( n+2 \right)}}$, where $A$ is a constant to be determined.

Hence find the exact value of $\sum\limits_{r=10}^{\infty }{\frac{1-r}{\left( r+1 \right)\left( r+2 \right)\left( r+3 \right)}}$.

2021 NYJC Promo Q5

<p>Given that&nbsp;${{u}_{n}}=\frac{{{2}^{n}}}{n}$, show that&nbsp;${{u}_{n+1}}-{{u}_{n}}=\frac{{{2}^{n}}\left( n-1 \right)}{n\left( n+1 \right)}$ where&nbsp;$n$ is a positive integer.</p>
<p>Hence, show that&nbsp;$\sum\limits_{n=1}^{N}{\frac{{{2}^{n}}\left( n-1 \right)}{n\left( n+1 \right)}}=\frac{{{2}^{N+1}}}{N+1}-2$.</p>

Given that ${{u}_{n}}=\frac{{{2}^{n}}}{n}$, show that ${{u}_{n+1}}-{{u}_{n}}=\frac{{{2}^{n}}\left( n-1 \right)}{n\left( n+1 \right)}$ where $n$ is a positive integer.

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

N1991 P1 Q3 (a)

<p>A sequence ${{u}_{1}},{{u}_{2}},{{u}_{3}},...$ is such that ${{u}_{n}}=\frac{1}{{{\left( 2n+1 \right)}^{2}}}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">By considering ${{u}_{n-1}}-{{u}_{n}}$, find $\sum\limits_{n=2}^{N}{\frac{n}{{{\left( 2n-1 \right)}^{2}}{{\left( 2n+1 \right)}^{2}}}}$.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">Give a reason why the series in part (i) is convergent and state the sum to infinity.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Use your answer to part (i) to find $\sum\limits_{n=3}^{N+2}{\frac{n-1}{{{\left( 2n-1 \right)}^{2}}{{\left( 2n-3 \right)}^{2}}}}$<span class="inline-marks">[3]</span></div></li></ol>

A sequence ${{u}_{1}},{{u}_{2}},{{u}_{3}},...$ is such that ${{u}_{n}}=\frac{1}{{{\left( 2n+1 \right)}^{2}}}$.

By considering ${{u}_{n-1}}-{{u}_{n}}$, find $\sum\limits_{n=2}^{N}{\frac{n}{{{\left( 2n-1 \right)}^{2}}{{\left( 2n+1 \right)}^{2}}}}$.

Give a reason why the series in part (i) is convergent and state the sum to infinity.

Use your answer to part (i) to find $\sum\limits_{n=3}^{N+2}{\frac{n-1}{{{\left( 2n-1 \right)}^{2}}{{\left( 2n-3 \right)}^{2}}}}$

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