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$$ \begin{array}{c|l} n & k \\ \hline 4&6\\ 5&9\\ 6&\frac{25}{2} \\ 7&\frac{33}{2} \\ 8&21 \\ 9&26 \\ 10&\frac{63}{2} \\ 11& \frac{75}{2} \end{array}$$

I observed the following pattern. $$\underbrace{6 \to 9}_{3} \underbrace{\to \frac{25}{2}}_{\frac{7}{2}} \underbrace{\to \frac{33}{2}}_{4} \underbrace{\to 21}_{\frac{9}{2}} \underbrace{\to 26}_{5} \underbrace{\to \frac{63}{2}}_{\frac{11}{2}} \underbrace{\to \frac{75}{2}}_{6}$$

Then the difference between two consecutive resulting numbers is $\frac{1}{2}$.

How to write the relation between $n$ & $k$?

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  • $\begingroup$ Where is this pattern coming from? Are you just given a bunch of terms? $\endgroup$ Apr 29, 2017 at 14:15
  • $\begingroup$ I found a pattern in this number set. I just $\endgroup$
    – Oily
    Apr 29, 2017 at 14:20
  • $\begingroup$ You can interpolate them with a function which has a derivative which integrates to a linear function. Which functions have a linear derivative? $\endgroup$ Apr 29, 2017 at 14:20
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    $\begingroup$ You've noticed that the second differences are constant, and so you should be looking for a quadratic equation. $\endgroup$
    – Joppy
    Apr 29, 2017 at 14:22
  • $\begingroup$ Hint: Write the integer differences as halves of integers and you'll see a clearer pattern. Then Use @Joppy 's observation to find a quadratic expression for $k$ as a function of $n$. $\endgroup$ Apr 29, 2017 at 14:27

3 Answers 3

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Hint Since the successive differences $\Delta k = k(n + 1) - k(n)$ are linear, $k$ is given by a quadratic expression in $n$.

Alternatively, one can modify appropriately the usual identity $$\sum_{i = 1}^m i = \frac{1}{2} m (m + 1)$$ to produce an expression for the partial sum of any arithmetic sequence.


In response to a comment, here's how one can apply the second hint:

The observed pattern says that $k(4) = 6$ and that the value $k(n)$ is given by adding to $k(4) = 6$ the $n - 4$ half-integers starting with $3$:

$$k(n) = 6 + \sum_{i = 1}^{n - 4} \left(3 + \frac{i - 1}{2}\right) .$$

Notice that if one only wants some explicit expression for $k(n)$, this above certainly suffices. It works for $n = 4$, too, taking the usual convention that an empty sum has value zero. Now, if one wants a closed-form expression for this quantity, we can manipulate this sum so that we can take advantage of the above formula for the sum of the first $m$ integers: $$ \begin{align*} k(n) &= 6 + \sum_{i = 1}^{n - 4} \left(\frac{5}{2} + \frac{i}{2}\right) \\ &= 6 + \frac{5}{2} (n - 4) + \frac{1}{2} \sum_{i = 1}^{n - 4} i \\ &= -4 + \frac{5}{2} n + \frac{1}{2} \sum_{i = 1}^{n - 4} i . \end{align*} $$ Taking $m = n - 4$ in the sum formula and substituting gives $$k(n) = -4 + \frac{5}{2} n + \frac{1}{2} \left[\frac{1}{2} (n - 4)[(n - 4) + 1]\right] = \frac{1}{4} n^2 + \frac{3}{4} n - 1.$$

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  • $\begingroup$ Thank you very much for answering ! $\endgroup$
    – Oily
    Apr 29, 2017 at 14:24
  • $\begingroup$ You're welcome. $\endgroup$ Apr 29, 2017 at 18:27
  • $\begingroup$ By using the recursive formula I could obtained the sequence. But I didn't understand the relation of the partial sum. $\endgroup$
    – Oily
    Apr 30, 2017 at 6:18
  • $\begingroup$ @Oily I've added some details about how one can use that formula for the partial sum to produce a closed-form formula for $k(n)$. $\endgroup$ Apr 30, 2017 at 10:20
  • $\begingroup$ You're welcome, I hope it was useful. $\endgroup$ Apr 30, 2017 at 12:12
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It is a sequence, right? You can define it recuresively as:

$a_0 = 0$, $a_1 = 1.5$ and $a_{n+1} = a_n + (a_n-a_{n-1}) + \frac{1}{2}$ for $n\geq 1$

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Hint$$g(k) = \int_{k}^{k+1} \frac{\partial f(x)}{\partial x} \partial x$$

where $g$ is the function you enumerate and the integral is your observation.

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