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So it is time to ask a question of my own, i do not need an entire solution, just what the next step is

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find the values of A and B
i did the first step which gives me

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not sure what to do next
@Callisto can you help?
you tagged someone with an "i hate math" picture to a math question?
Is it n(2n^2 + 3n -2) on the right side?
yeah it is
|dw:1350056495046:dw| Can you simplify it first?
which side? left or right
Both. There is a common factor.
that should be -2n on ther ight
Not really.\[\frac{A}{6}n(n+1)(2n+1) +bn= n(n^2 + 3n -2)\]\[n[\frac{A}{6}(n+1)(2n+1)+b] = n(n^2 + 3n -2)\] Cancel the common factor. and expand the left side.
you could change b to 6b/6 and take out a factor of 1/6 right?
You can, of course. But there is a common factor on both sides that you can cancel, what is it?
get rid of n on both sides? if not then i dont know waht the common factor is on both sides
Yes! So, now, you get \[\frac{A}{6}(n+1)(2n+1)+b = n^2 + 3n -2\]So, expand the terms on the left. What do you get?
now i shall multiply out
forgot my A's
Not really..\[\frac{A}{6}(2n+1)(n+1) = \frac{A}{6}(2n^2+3n+1) = ...?\]
oh and it is 2n² on the right
so that means to get 2n² A has to be 6
I'm sorry but I have to go now... Basically, you need to expand the first term on the left, combine the like terms and compare the coefficient of x^2, x and constant terms on both sides. Then you can find A and B. As for your expansion, sorry again that it is wrong.
okay thank you! i shall work on this thank you for your help
a has to be 6 is correct, I supposed :|
Once you find a, then compare the constant term on both sides. You should be able to get b. Good luck :)
b is-3
thank you :D

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