There isn't a particular formula.

Rather, the question expects you to discover that :

2 = 2 x 1
6 = 3 x 2
12 = 4 x 3
20 = 5 x 4
30 = 6 x 5
42 = 7 x 6
56 = 8 x 7

Well what if let’s say the question were to go something like: 8, 19 (+11), 33 (+14), 50 (+17), 69 (+19), etc....

How would that work out?

8 = 4*2
19 = 19*1
33= 3*11

???

Is that an actual question given? or you came up with the numbers yourself?

When such questions are set in the exam, the pattern has to be formed first by the setter. He/she will then put in numbers that satisfy the pattern

If you meant to type 70 instead of 69 , the pattern for 8,19,33,50,70... would be

½(3n² + 13n)


Another example :

50,55(+5),63(+8),74(+11),88(+14),105(+17),..


The pattern here would be

½(3n² + n + 96)


And yet another one :

23,59(+36),98(+39),140(+42),185(+45),...

The pattern here would be

½(3n² + 63n - 20)


As you can see there's no fixed pattern even though they all increase by more than 3 of the previous increase.

It depends on the number you start with and what the first increment is.

Thanks for the reply! Seeing the examples you have listed, each of the answers have :1/2(3n2). Where do you get 1/2 from? Where do you get 3n2 from? What if the difference between each term were to increase by 4 each time. Will the answer be along the lines of 1/2(4n2)?

For 8,19,33,50,70 :

First term :
8 = 5 + 3
= 1 x 5 + 3 x 1

2nd term :
19 = 8 + 11
= 5 + 3 + 5 + 3 x 2
= 2 x 5 + 3 x (1 + 2)

3rd term :
33 = 19 + 14
= 2 x 5 + 3 + (1 + 2) + 5 + 3 x 3
= 3 x 5 + 3 x (1 + 2 + 3)


4th term :
50 = 33 + 17
= 3 x 5 + 3 x (1 + 2 + 3) + 5 + 3 x 4
= 4 x 5 + 3 x (1 + 2 + 3 + 4)


5th term :
70 = 50 + 20
= 4 x 5 + 3 x (1 + 2 + 3 + 4) + 5 + 3 x 5
= 5 x 5 + 3 x (1 + 2 + 3 + 4 + 5)


So for the nth term,

nth term = n x 5 + 3 x (1 + 2 + 3 +... + (n-2) + (n - 1) + n)

Now for the series

(1 + 2 + 3 +... + (n-2) + (n - 1) + n) ,

Notice that you can pair the numbers
1 + n = n + 1
2 + (n-1) = n + 1
3 + (n-2) = n + 1
And so on.

Each pair's total is n + 1. The number of pairs is ½ of the total number of terms in the series (which is n) , so it equals ½n.

so the sum = number of pairs x total of each pair

=½n(n + 1)


Therefore

nth term = n x 5 + 3 x (1 + 2 + 3 +... + (n-2) + (n - 1) + n)

= 5n + 3 x (½n(n + 1))

= 5n + 3/2 n² + 3/2 n

= 3/2 n² + 13/2 n

= ½(3n² + 13n)

You've kind of spotted a trend.

If the difference increases by u each time, then the first part of the general term to be found could be said to be

u/2 n²

But you will still need to figure out the other parts of the term

Ok let's modify the sequence a little such that any increase is always 4 more than the previous increase.

8,19,34,53,76

The pattern is : 2n² + 5n + 1


Modify the previous 2nd example,

50,55(+5),64(+9),77(+13),94(+17),115(+21),..

Pattern becomes : 2n² - n + 49


Modify the 3rd last example,

23,59(+36),99(+40),143(+44),191(+48),...


Pattern becomes: 2n² + 30n - 9



Now 2n² = 4/2 n²


You can try on your own for other difference increases. But the difficult part is to derive the rest of the expression as it would involve some trial and error.


I would recommend that you first try to spot a pattern, then work towards the obtaining the final expression
(something like my example for ½(3n² + 13n) .

Thank you for all this!!! I’d like to ask another question.

8,19,33,50,70

8= 5+3
...

19= 8+11
...

33= 19+ 14

How do you pick the numbers for the first term (why specifically
8= 5+3, and not 8= 4+4?)?I managed to see why you picked the specific numbers for the second and 3rd terms, but not the first.

I also tried doing this for when the difference in the numbers +4 each time

8, 19, 34, 53, 76

(The first pairing is probably wrong)

First term
8= 4+4
=1*4 + 4*1

Second term
19= 8+ 11
4*2 + ???

Because the difference increases by 3 each time. It's more of rearranging the terms or whichever pattern you see first.


You could also do it like this :
For 8,19,33,50,70 :

First term :
8

2nd term :
19 = 8 + 11
= 8 + 8 + 3
= 8 x 2 + 3 x 1

3rd term :
33 = 19 + 14
= 8 x 2 + 3 x 1 + 8 + 3 x 2
= 8 x 3 + 3 x (1 + 2)


4th term :
50 = 33 + 17
= 8 x 3 + 3 x (1 + 2) + 8 + 3 x 3
= 8 x 4 + 3 x (1 + 2 + 3)


5th term :
70 = 50 + 20
= 8 x 4 + 3 x (1 + 2 + 3) + 8 + 3 x 4
= 8 x 5 + 3 x (1 + 2 + 3 + 4)


So for the nth term,

nth term = 8 x n + 3 x (1 + 2 + 3 +... + (n-2) + (n - 1) )

Now for the series

(1 + 2 + 3 +... + (n-2) + (n - 1) + n) ,


Sum = ½(n -1)(n - 1 + 1)
= ½(n - 1)(n)
= ½n² - ½n


Therefore,

nth term = 8 x n + 3(½n² - ½n)

= 8n + 3/2 n² - 3/2 n

3/2 n² + 13/2 n

= ½ (3n² + 13n)


There's no need to specifically pick the numbers 5 + 3 , it all depends on what insight/discovery you make of the sequence first.


As for why not 4 + 4, it's because there is no relation to the increase of 3 for every difference. Splitting 8 into 5 + 3 means another 3 is created , which can be grouped with the other 3s in the sum

Now for 8,19,34,53,76 :


First term :
8 = 1 + 7
= 1 + 4 + 3
= 1 + 4 x 1 + 3 x 1


2nd term :
19 = 8 + 11
= 1 + 4 x 1 + 3 + 4 x 2 + 3
= 1 + 4 x (1 + 2) + 3 x 2

3rd term :
34 = 19 + 15
= 1 + 4 x 3 + 3 x 2 + 4 x 3 + 3
= 1 + 4 x (1 + 2 + 3) + 3 x 3


4th term :
53 = 34 + 19
= 1 + 4 x (1 + 2 + 3) + 3 x 3 + 4 x 4 + 3
= 1 + 4 x (1 + 2 + 3 + 4) + 3 x 4

5th term :
76 = 53 + 23
= 1 + 4 x (1 + 2 + 3 + 4) + 3 x 4 + 4 x 5 + 3
= 1 + 4 x (1 + 2 + 3 + 4 + 5) + 3 x 5


So for the nth term,

nth term
1 + 4 x (1 + 2 + 3 +... + (n-2) + (n-1) + n) + 3 x n

= 1 + 4( ½n(n+1) ) + 3n

= 1 + 2n² + 2n + 3n

= 2n² + 5n + 1

Thank you soooo much!!!!!! I understand it now.

Welcome

Here's an alternative for 8,19,34,53,76 :


First term :
8 = 1 + 7

2nd term :
19 = 8 + 11
= 1 + 7 + 7 + 4
= 1 + 7 x 2 + 4 x 1

3rd term :
34 = 19 + 15
= 1 + 7 x 2 + 4 x 1 + 7 + 4 x 2
= 1 + 7 x 3 + 4 x (1 + 2)


4th term :
53 = 34 + 19
= 1 + 7 x 3 + 4 x (1 + 2) + 7 + 4 x 3
= 1 + 7 x 4 + 4 x (1 + 2 + 3)

5th term :
76 = 53 + 23
= 1 + 7 x 4 + 4 x (1 + 2 + 3) + 7 + 4 x 4
= 1 + 7 x 5 + 4 x (1 + 2 + 3 + 4)


So for the nth term,

nth term
1 + 7 x n + 4 x (1 + 2 + 3 +... + (n-2) + (n-1) )

= 1 + 7n + 4(½(n - 1)(n)

= 1 + 7n + 2n² - 2n

= 2n² + 5n + 1