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<p>The diagram shows the graph of $y={{2}^{x}}-3x$. The two roots of the equation ${{2}^{x}}-3x=0$ are denoted by $\alpha $ and $\beta $, where $\alpha <\beta $. A sequence of real numbers ${{x}_{1}}$, ${{x}_{2}}$, ${{x}_{3}}$, … satisfies the recurrence relation ${{x}_{n\,+\,1}}=\frac{1}{3}\left( {{2}^{x{{\,}_{n}}}} \right)$ for $n\ge 1$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Prove algebraically that, if the sequence converges, then it converges to either $\alpha $ or $\beta $.</div></li><li><div class="list-item-paragraph">For the case ${{x}_{1}}=4$, describe the behaviour of the sequence using a calculator.</div></li><li><div class="list-item-paragraph">By considering ${{x}_{n+1}}-{{x}_{n}}$, prove that ${{x}_{n+1}}>{{x}_{n}}$ if ${{x}_{n}}>\beta $. Hence, justify your answer in part <strong>(ii)</strong>.</div></li></ol>

The diagram shows the graph of $y={{2}^{x}}-3x$. The two roots of the equation ${{2}^{x}}-3x=0$ are denoted by $\alpha $ and $\beta $, where $\alpha <\beta $. A sequence of real numbers ${{x}_{1}}$, ${{x}_{2}}$, ${{x}_{3}}$, … satisfies the recurrence relation ${{x}_{n\,+\,1}}=\frac{1}{3}\left( {{2}^{x{{\,}_{n}}}} \right)$ for $n\ge 1$.

Prove algebraically that, if the sequence converges, then it converges to either $\alpha $ or $\beta $.

For the case ${{x}_{1}}=4$, describe the behaviour of the sequence using a calculator.

By considering ${{x}_{n+1}}-{{x}_{n}}$, prove that ${{x}_{n+1}}>{{x}_{n}}$ if ${{x}_{n}}>\beta $. Hence, justify your answer in part <strong>(ii)</strong>.

Limit of a Sequence and Behaviour of a Sequence Q2

Limit of a Sequence and Behaviour of a Sequence Q2

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