<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>

Show that $\frac{4}{\left( r-1 \right)!}-\frac{1}{r!}-\frac{3}{\left( r+1 \right)!}=\frac{4{{r}^{2}}+3r-4}{\left( r+1 \right)!}$ .

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 $n$.

Deduce that $\sum\limits_{r=1}^{n}{\frac{{{r}^{2}}-1}{\left( r+1 \right)!}<\frac{7}{4}}$ .

Method of Difference

Polynomials, Cubic Equations and Partial Fractions

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

2018 EJC PROMO Q8

2018 EJC PROMO Q8

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