Question

Describe the pattern
2,6,12,20,30 Describe the pattern
2,6,12,20,30 @Mathematics

Answer 1

work out the difference between each term - that should give you a clue...

Answer 2

I understand that , ive been having some trouble with sequences in class.

Answer 3

so what do you get when you work out the difference between each term in this case?

Answer 4

where are you getting stuck?

Answer 5

determining the pattern, the and finding the rule for the nth term

Answer 6

you need to break this up into steps.
1. first write down the difference between each pair of terms on a piece of paper.
2. then look at this new sequence and see if you can spot a pattern.
3. lastly try and use this to work out the equation for the nth term.
so, why don't you start with steps 1 and 2 - what do you get?

Answer 7

have you learnt about arithmetic progressions/sequences?

Answer 8

the pattern looks like 4,6,8, and 10

Answer 9

ok - so now you know the difference between each term is an increasing sequence of even numbers

Answer 10

so, we can write the sequence like this:
2, 2+4, 2+4+6, 2+4+6+8, 2+4+6+8+10
does that help?

Answer 11

yeah so

Answer 12

each term is the sum of an arithmetic progression. do you know the formulae for the sum of an arithmetic progression?

Answer 13

is it an=a1+ (n-1)d ?

Answer 14

yes, so the 1st term is just the sum up to n=1
the 2nd term is the sum up to n=2
etc

do you think you can work this out now?

Answer 15

wait so the common difference would be written 4,6,8,10?

Answer 16

look at the 4th term, it is simply: 2+4+6+8
do you know the formulae for the SUM of an arithmetic sequence?

Answer 17

\[S_n=\frac{n}{2}(2a_1+(n-1)d)\]

Answer 18

in your case:\[n=4, a_1=2, d=2\]

Answer 19

similarly, the 3rd term is just the SUM of the 1st 3 terms
the 2nd term is the SUM of the 1st 2 terms

so the nth term would be just the SUM of the 1st n terms

Answer 20

so what would be the rule , how do i find that

Answer 21

I gave you the formulae above, it is:\[S_n=\frac{n}{2}(2a_1+(n-1)d)\]
where, for your particular sequence:\[a_1=2, d=2\]

Answer 22

so for your sequence the nth term is given by:\[a_n=\frac{n}{2}(2*2+(n-1)*2)\]\[=\frac{n}{2}(4+2n-2)=\frac{n}{2}(2+2n)=n(n+1)\]
do you see how we got to this?

Answer 23

yeah i see now

Answer 24

these can be tricky, you just need to be patient and persevere...