2022 HCI J1 BT Q7
It is given that ${{u}_{r}}=\frac{1}{\left( r+2k \right)!}$, where $r\in {{\mathbb{Z}}^{+}}$ and $k$ is a positive constant.
(i)
Show that ${{u}_{r}}-{{u}_{r+1}}=\frac{r+2k}{\left( r+2k+1 \right)!}$.
[1]
(ii)
Hence use the method of differences to find $\sum\limits_{r=1}^{n}{\frac{r+2k}{\left( r+2k+1 \right)!}}$ in terms of $n$ and $k$. You need not simplify your answer.
[3]
Let ${{S}_{n}}$ denote the sum of the first $n$ terms of the series
$\frac{1}{4\,\,\,2!}+\frac{1}{5\,\,\,3!}+\frac{1}{6\,\,\,4!}+…$.
(iii)
By finding a suitable integer value of k and using the result obtained in part (ii), find ${{S}_{n}}$ in terms of $n$ .
[2]
(iv)
Deduce that ${{S}_{n}}<\frac{1}{6}$ for all $n\in {{\mathbb{Z}}^{+}}$.
[1]
(v)
Find the least value of $n$ for which ${{S}_{n}}$ is within ${{10}^{-20}}$ of the sum to infinity.
[2]
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