An arithmetic progression $\left\{ {{u}_{n}} \right\}$ is grouped into sets containing $1$, \(3\

Question

An arithmetic progression $\left\{ {{u}_{n}} \right\}$ is grouped into sets containing $1$, $3$, $9$, … shown below, so that the number of integers in the $n$$^{\mathrm{th}}$ set is ${{3}^{n-1}}$. $\left\{ 1 \right\}$, $\left\{ 5,\mathrm{ }9,\mathrm{ }13 \right\}$, $\left\{ 17,\mathrm{ }21,\mathrm{ }25,\mathrm{ }29,\mathrm{ }33,\mathrm{ }37,\mathrm{ }41,\mathrm{ }45,\mathrm{ }49 \right\}$, … In either order, find expressions in terms of $n$, for the first integer in the $n$$^{\mathrm{th}}$ set,[3] the last integer in the $n$$^{\mathrm{th}}$ set.[2] Show that the sum of the terms in the $n$$^{\mathrm{th}}$ set is given by ${{3}^{n}}\left[ A\left( {{3}^{n-2}} \right)+B \right]$, where $A$ and $B$ are integers to be determined.[4]

Accepted Answer

(i)(a) $2\cdot3^{n-1}-1$ (i)(b) $2\cdot3^n-5$ (ii) $A=4$, $B=-1$

2007 HCI P1 Q9

2007 HCI P1 Q9

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