// StartMathJax Script window.MathJax = {loader: {load: [ 'input/asciimath', 'ui/lazy', 'output/chtml', 'ui/menu']} }; (function() { var script = document.createElement('script'); script.src = "https://cdn.jsdelivr.net/npm/mathjax@3/es5/startup.js"; script.async = true; document.head.appendChild(script); })(); ---------- (Different files) ---------- // UpdateTypeset Script config = { attributes: true, childList: true, subtree: true }; // Callback function to execute when mutations are observed callback = (mutationList, observer) => { for (mutation of mutationList) { if (mutation.type === 'childList') { console.log('A child node has been added or removed.'); MathJax.typeset(); } else if (mutation.type === 'attributes') { console.log(`The ${mutation.attributeName} attribute was modified.`); } } }; // Create an observer instance linked to the callback function observer = new MutationObserver(callback); document.onreadystatechange = () => { if (document.readyState === 'complete') { console.log("Loaded fully according to readyState") targetNode = document.getElementById('content-wrapper') console.log(targetNode) // Start observing the target node for configured mutations observer.observe(targetNode, config); } }
top of page

How to Handle Quadratic Sequences

Updated: Nov 27, 2022

There are several ways to find the equation for a quadratic sequence - I have found the following to be the best.


Quadratic sequences take the form:


If we substitute the first term (1) into the general form, we get…

Which simplifies to…

This is shown in the table below immediately beneath the term’s value.

Here we have a sequence of numbers. I have included the differences between the numbers in the sequence.

Notice that the 2nd difference is constant across the sequence- a sure sign that you have a quadratic sequence.


Finding the next number in the sequence is easy - add the 1st and second differences together from n=4 to n=5 (58 + 8 = 66) then add this to the value for n=5 (207 + 66 = 273). The value for n=6 is 273.


Let's find out the quadratic equation that gave this sequence of numbers...


From the derived equations in the first table, we can calculate the values for a, b and c.


Sequence = a + b + c = 23


1st difference = 3a + b = 34

2nd difference = 2a = 8


Of course, you can memorise the formulas, but they can still be worked out if you forget.


For a:

2a = 8

So a = 4


Substitute this value for a in 3a + b = 34

3(4) + b = 34

So b = 22


Finally…

a + b + c = 23

4 + 22 + c = 23

c = -3


We can put these values into the quadratic equation and, voila, get...

Make sure you check it with a value for n, for example n = 2... we should get 57, the second number in the sequence.



What if we need to find a value in the sequence, perhaps the number for when n = 20? Just substitute into the equation...



When n = 20, the number in the sequence is 2037.


Until the next time... as always, comments and thoughts are welcome.

 



30 views0 comments

Recent Posts

See All

Comentários

Avaliado com 0 de 5 estrelas.
Ainda sem avaliações

Adicione uma avaliação
bottom of page