By considering ${{u}_{k}}-{{u}_{k+1}}$, where ${{u}_{k}}=\frac{1}{k!}$, find \(\frac{3}{4!}

Question

By considering ${{u}_{k}}-{{u}_{k+1}}$, where ${{u}_{k}}=\frac{1}{k!}$, find $\frac{3}{4!}+\frac{4}{5!}+\frac{5}{6!}+...+\frac{3n+2}{\left( 3n+3 \right)!}$ in terms of $n$.[4] Find $\sum\limits_{r=5}^{3n+3}{\frac{r-1}{r!}}$. Hence show that $\sum\limits_{r=5}^{3n+3}{\frac{3}{r!}<\frac{1}{24}}$.[4]

Accepted Answer

(i) $\frac{1}{6}-\frac{1}{\left( 3n+3 \right)!}$ (ii) $\frac{1}{24}-\frac{1}{\left( 3n+3 \right)!}$

2022 NJC P1 Q6

2022 NJC P1 Q6

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