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Mathematics 14 Online
OpenStudy (anonymous):

write down an expression, in terms of n, for the nth term formula of the sequence? 0,3,8,15,_____

OpenStudy (anonymous):

@experimentX @dpaInc @ParthKohli

OpenStudy (anonymous):

its not common?

OpenStudy (experimentx):

0,3,8,15, ... every term is of 3+2(n) more than the previous term

OpenStudy (anonymous):

bro, how do u get it? are there any methods? please try to help?

OpenStudy (anonymous):

where do iget the general formula for this?

OpenStudy (anonymous):

It would be (n^2+3n-4)/2.

OpenStudy (anonymous):

how do u get it/

OpenStudy (experimentx):

should be n(n+2)

OpenStudy (anonymous):

Use the second difference formula: a+d(n-1)+1/2(n-1)(n-2)c where a=first term, d=difference of first two terms c=second common difference

OpenStudy (anonymous):

wow...i never came across this...where did u get this?

OpenStudy (anonymous):

The internet! :D I forgot the website though.

OpenStudy (anonymous):

how do u simplify this 1/2(n-1)(n-2)c?

OpenStudy (anonymous):

(n^2-3n+c)/2

OpenStudy (anonymous):

Do not replace 'n'. Just put 2 in place of c. Since it is the second common difference. 0--3 is 3| 3--8 is 5| The difference of 3 and 5 is 2.

OpenStudy (anonymous):

do u have a webste for this, where did u search it?

OpenStudy (anonymous):

I lost the link to the website I found it. But you can go through this one: http://www.skoool.co.za/studynotes/maths/id635.htm It has a few examples.

OpenStudy (experimentx):

Oh ... i didn't know they were called quadratic sequence and have formula ...

OpenStudy (anonymous):

HAHA! You know it now. No worries! :D

OpenStudy (anonymous):

it is (n^2-1) thanks alot @emcrazy14 for your method...:DDDD, and @experimentX u too for trying to help me:)

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