# Help with Induction Questions (1 Viewer)

#### policy

##### New Member
1) Prove, for all positive integers $\bg_white n$, the identity
$\bg_white \frac{1}{n+1}+ \frac{1}{n+2} +... +\frac{1}{2n} =1- \frac{1}{2}+ \frac{1}{3}-...+ \frac{1}{2n-1} - \frac{1}{2n}.$

2) The sequence $\bg_white x_{n}$ is given by $\bg_white x_{1} = 1$ and $\bg_white x_{n+1} = \frac{4+ x_{n} }{1+ x_{n} }$ for $\bg_white n \geq 1.$

a) Prove by induction that for $\bg_white n \geq 1.$, $\bg_white x_{n} = 2 \big( \frac{1+ \alpha ^{n} }{1- \alpha ^{n}} \big)$, where $\bg_white \alpha = \frac{-1}{3}.$

b) Hence find the limiting value of $\bg_white x_{n}$ as $\bg_white n \rightarrow \infty.$

3) A sequence is defined by $\bg_white a_{n+1} = \frac{1}{2} \big( a_{n}+ \frac{2}{ a_{n} } \big)$ where $\bg_white a_{1} =1$ and $\bg_white n$ is a positive integer.

a) Use induction to show that $\bg_white \frac{ a_{n} - \sqrt{2} }{a_{n} + \sqrt{2}} = \big( \frac{1- \sqrt{2} }{1+ \sqrt{2} } \big)^ \big(2^{n-1}\big)$ (the 2^n-1 is the power of the adjacent of fraction)

b) Hence find the limiting value of $\bg_white x_{n}$ as $\bg_white n$ becomes large.

Any help would greatly be appreciated!

#### Drongoski

##### Well-Known Member
1) Prove, for all positive integers $\bg_white n$, the identity
$\bg_white \frac{1}{n+1}+ \frac{1}{n+2} +... +\frac{1}{2n} =1- \frac{1}{2}+ \frac{1}{3}-...+ \frac{1}{2n-1} - \frac{1}{2n}.$

2) The sequence $\bg_white x_{n}$ is given by $\bg_white x_{1} = 1$ and $\bg_white x_{n+1} = \frac{4+ x_{n} }{1+ x_{n} }$ for $\bg_white n \geq 1.$

a) Prove by induction that for $\bg_white n \geq 1.$, $\bg_white x_{n} = 2 \big( \frac{1+ \alpha ^{n} }{1- \alpha ^{n}} \big)$, where $\bg_white \alpha = \frac{-1}{3}.$

b) Hence find the limiting value of $\bg_white x_{n}$ as $\bg_white n \rightarrow \infty.$

3) A sequence is defined by $\bg_white a_{n+1} = \frac{1}{2} \big( a_{n}+ \frac{2}{ a_{n} } \big)$ where $\bg_white a_{1} =1$ and $\bg_white n$ is a positive integer.

a) Use induction to show that $\bg_white \frac{ a_{n} - \sqrt{2} }{a_{n} + \sqrt{2}} = \big( \frac{1- \sqrt{2} }{1+ \sqrt{2} } \big)^ \big(2^{n-1}\big)$ (the 2^n-1 is the power of the adjacent of fraction)

b) Hence find the limiting value of $\bg_white x_{n}$ as $\bg_white n$ becomes large.

Any help would greatly be appreciated!
Q2

Very heavy LaTeX typing! I'll skip the many little steps that you can easily do yourself.

a) OK - you can show true for n=1.

Let formula hold for n = k >=1

$\bg_white i.e. x_n = 2 \left ( \frac {1+\alpha ^k}{1 - \alpha ^k} \right ) \\ \\ \therefore x_{n+1} = \frac {4 + 2(\frac {1+\alpha ^k}{1 - \alpha ^k})}{1 + 2(\frac {1 + \alpha ^k}{1 - \alpha ^k})} = \frac {4(1-\alpha^k) + 2(1+\alpha^k)}{(1-\alpha^k) + 2(1+\alpha^k)} = \frac {6 - 2\alpha^k}{3 + \alpha^k} = \frac {6 + 6 \times (-\frac {1}{3})\times \alpha^k}{3 - 3 \times (-\frac {1}{3}) \times \alpha^k}\\ \\ = \frac {6 + 6 \times \alpha \times \alpha^k}{3 - 3 \times \alpha \times \alpha ^k} = \frac {6(1 + \alpha ^{k+1})}{3(1 - \alpha ^ {k+1})} = 2\left ( \frac {1 + \alpha ^{k+1}}{1 - \alpha ^{k+1}}\right )$

.: if true for n = k, true also for n = k+1

So you have essentially proven the formula.

b) $\bg_white as n \rightarrow \infty x_n \rightarrow 2\left ( \frac {1 + 0}{1 - 0} \right ) = 2.$

I will only do this bit. Hope it helps.

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