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Let the sequence $\left( x_n \right)_{n \in \mathbb{N}}$ of real numbers be defined inductively as follows: $$ x_1 \colon= 1, \ x_2 \colon= 2, \ \mbox{ and } \ x_n \colon= \frac{1}{2} \left( x_{n-1} + x_{n-2} \right) \ \mbox{ for any natural number } n > 2. $$

Then how to find an explicit, non-inductive formula for $x_n$?

And, how to show the following? $$ x_{2n+1} = 1 + \frac{1}{2} + \frac{1}{2^3} + \cdots + \frac{1}{2^{2n-1}} $$ for all $n \in \mathbb{N}$.

My Attempt:

We note that $x_1 = 1$, $x_2 = 2$, and so $$ x_3 = \frac{1}{2} \left( x_2 + x_1 \right) = \frac{3}{2} = 1 + \frac{1}{2} = 1 + \frac{1}{2^1}, $$ $$ x_4 = \frac{1}{2} \left( x_3 + x_2 \right) = \frac{7}{4} = 1 + \frac{1}{2} + \frac{1}{4} = 1 + \frac{1}{2^1} + \frac{1}{2^2}, $$ $$ x_5 = \frac{1}{2} \left( x_4 + x_3 \right) = \frac{13}{8} = 1 + \frac{1}{2} + \frac{1}{8} = 1 + \frac{1}{2} + \frac{1}{2^3}, $$ $$ x_6 = \frac{1}{2} \left( x_5 + x_4 \right) = \frac{27}{16} = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} = 1 + \frac{1}{2^1} + \frac{1}{2^3} + \frac{1}{2^4}, $$ $$ x_7 = \frac{1}{2} \left( x_6 + x_5 \right) = \frac{53}{32} = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{32} = 1 + \frac{1}{2^1} + \frac{1}{2^3} + \frac{1}{2^5}, $$ $$ x_8 = \frac{1}{2} \left( x_7 + x_6 \right) = \frac{107}{64} = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \frac{1}{64} = 1 + \frac{1}{2^1} + \frac{1}{2^3} + \frac{1}{2^5} + \frac{1}{2^6}, $$ $$ x_9 = \frac{1}{2} \left( x_8 + x_7 \right) = \frac{213}{128} = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \frac{1}{128} = 1 + \frac{1}{2^1} + \frac{1}{2^3} + \frac{1}{2^5} + \frac{1}{2^7}, $$ $$ x_{10} = \frac{1}{2} \left( x_9 + x_8 \right) = \frac{427}{256} = 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \frac{1}{256} = 1 + \frac{1}{2^1} + \frac{1}{2^3} + \frac{1}{2^5} + \frac{1}{2^7} + \frac{1}{2^8}. $$ So, for each $n \in \mathbb{N}$, we have $$ x_{2n} = 1 + \frac{1}{2^1} + \frac{1}{2^3} + \frac{1}{2^5} + \cdots + \frac{1}{2^{2n-3}} + \frac{1}{2^{2n-2}} = 1 + \frac{1}{2} \left( 1 + \frac{1}{4} + \frac{1}{4^2} + \cdots + \frac{1}{4^{n-2}} \right) + \frac{1}{4^{n-1}} = 1 + \frac{1}{2} \frac{ 1 - \left( \frac{1}{4} \right)^{n-1} }{ 1 - \frac{1}{4} } + \frac{1}{4^{n-1}} = 1 + \frac{2}{3} \left( 1 - \left( \frac{1}{4} \right)^{n-1} \right) + \frac{1}{4^{n-1}} = \frac{1}{3} \left( 5 + \frac{1}{4^{n-1} } \right) = \frac{1}{3} \left( 5 + \frac{ 4 }{ 4^n } \right), $$ and $$ x_{2n+1} = 1 + \frac{1}{2^1} + \frac{1}{2^3} + \frac{1}{2^5} + \ldots + \frac{1}{2^{2n-1}} = 1 + \frac{1}{2} \left( 1 + \frac{1}{4} + \frac{1}{4^2} + \cdots + \frac{1}{4^{n-1} } \right) = 1 + \frac{ \frac{1}{2} \left( 1 - \left( \frac{1}{4} \right)^n \right) }{ 1 - \frac{1}{4} } = 1 + \frac{2}{3} \left( 1 - \left( \frac{1}{4} \right)^n \right) = \frac{1}{3} \left( 5 - \frac{2}{4^n} \right). $$

Thus we have obtained $$ x_{2n} = \frac{1}{3} \left( 5 + \frac{ 4 }{ 4^n } \right), \tag{1} $$ and $$ x_{2n+1} = \frac{1}{3} \left( 5 - \frac{2}{4^n} \right) \tag{2} $$ for each $n \in \mathbb{N}$.

Now suppose that, for some $n \in \mathbb{N}$, the formulas (1) and (2) hold. Then we find that $$ x_{2n+2} = \frac{1}{2} \left( x_{2n+1} + x_{2n} \right) = \frac{1}{2} \left( \frac{1}{3} \left( 5 - \frac{2}{4^n} \right) + \frac{1}{3} \left( 5 + \frac{ 4 }{ 4^n } \right) \right) = \frac{1}{3} \left( 5 + \frac{1}{4^n} \right) = \frac{1}{3} \left( 5 + \frac{4}{4^{n+1}} \right), \tag{3} $$ and then $$ x_{2n+3} = \frac{1}{2} \left( x_{2n+2} + x_{2n+1} \right) = \frac{1}{2} \left( \frac{1}{3} \left( 5 + \frac{4}{4^{n+1}} \right) + \frac{1}{3} \left( 5 - \frac{2}{4^n} \right) \right) = \frac{1}{3} \left(5 - \frac{2}{4^{n+1} } \right). \tag{4} $$

Thus we find that formulas (3) and (4) are just the formulas (1) and (2), respectively, with $n$ replaced by $n+1$. This establishes that formulas (1) and (2) hold for all $n \in \mathbb{N}$.

Is there anything wrong with this reasoning?

Saaqib Mahmood
  • 26,762

3 Answers3

1

$$ x_{n+2}=\frac{1}{2}(x_{n+1}+x_{n}) \quad\Longrightarrow\quad x_{n+2}-x_{n+1}=-\frac{1}{2}(x_{n+1}-x_n). $$ So, if we set $y_n=x_{n+1}-x_n$, then the sequence $\{y_n\}$ satisfies $$ y_{n}=-\frac{1}{2}y_{n-1}=\left(-\frac{1}{2}\right)^2y_{n-2}=\cdots= \left(-\frac{1}{2}\right)^{n-1}y_1=\left(-\frac{1}{2}\right)^{n-1}(x_2-x_1)=\left(-\frac{1}{2}\right)^{n-1}, $$ and hence $$ x_{n+1}-x_n=\left(-\frac{1}{2}\right)^{n-1}, $$ and thus $$ x_n=(x_n-x_{n-1})+(x_{n-1}-x_{n-2})+\cdots+(x_2-x_1)+x_1=\left(-\frac{1}{2}\right)^{n-2}+\left(-\frac{1}{2}\right)^{n-3}+\cdots+\left(-\frac{1}{2}\right)^{1}+1+1 \\=\frac{\left(-\frac{1}{2}\right)^{n-1}-1}{-\frac{1}{2}-1}+1=1+\frac{2}{3}\left(1-\left(-\frac{1}{2}\right)^{n-1}\right) $$ Therefore $$ x_{2n+1}=1+\frac{2}{3}\left(1-\frac{1}{2^{2n}}\right)=1+\frac{1}{2}\cdot\frac{1-\frac{1}{2^{2n}}}{1-\frac{1}{2^2}}=1+\frac{1}{2}\left(1+\frac{1}{2^2}++\frac{1}{2^4}+\cdots+\frac{1}{2^{2n-2}}\right)\\=1+\frac{1}{2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{2n-1}} $$

Yiorgos S. Smyrlis
  • 83,933
0

Then how to find an explicit, non-inductive formula for $x_n$?

Hint: $\; x_n -x_{n-1}= \frac{-1}{2} \left( x_{n-1} - x_{n-2} \right) = \left(\frac{-1}{2}\right)^2\left( x_{n-2} - x_{n-3} \right) = \ldots\;$, so $\,x_n-x_{n-1}\,$ is a geometric progression, then $\,x_n\,$ is its sum plus some constant.

dxiv
  • 76,497
0

Note that $2x_n+x_{n-1}=2x_{n-1}+x_{n-2}$ for all $n\ge3$.

So, we have $2x_n+x_{n-1}=2x_{2}+x_{1}=5$ for all $n\ge 2$.

\begin{align*} -2x_n&=x_{n=1}-5\\ (-2)^nx_n&=(-2)^{n-1}x_{n-1}-5(-2)^{n-1}\\ \sum_{k=2}^n(-2)^kx_k&=\sum_{k=2}^n(-2)^{k-1}x_{k-1}-5\sum_{k=2}^n(-2)^{k-1}\\ (-2)^nx_n&=-2x_1-5\sum_{k=2}^n(-2)^{k-1}\\ (-2)^nx_n&=-2-5\times\frac{-2[(-2)^{n-1}-1]}{-2-1}\\ (-2)^nx_n&=\frac{4}{3}-\frac{10}{3}(-2)^{n-1}\\ x_n&=\frac{5}{3}+\frac{4}{3}(-2)^{-n} \end{align*}

Note that $\displaystyle x_k-x_{k-1}=\frac{-1}{2}(x_{k-1}-x_{k-2})$ for $k\ge3$

So, $\displaystyle x_k-x_{k-1}=\left(\frac{-1}{2}\right)^{k-2}(x_2-x_1)=\left(\frac{-1}{2}\right)^{k-2}$ for $k\ge 2$

$\displaystyle x_{2k+1}-x_{2k-1}=(x_{2k+1}-x_{2k})+(x_{2k}-x_{2k-1})=\left(\frac{-1}{2}\right)^{2k-1}+\left(\frac{-1}{2}\right)^{2k-2}=\left(\frac{1}{2}\right)^{2k-1}$

$$x_{2n+1}=\sum_{k=2}^n(x_{2k+1}-x_{2k-1})+x_1=1+\frac{1}{2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{2n-1}}$$

CY Aries
  • 23,393