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

2017 PJC MYCT P1 Q2

2017 PJC MYCT P1 Q2

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