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did you see my earlier reply to this?
i dont see it :(
oh wait nvm ignore the early one do you know anything about induction?
some1 tell me put n=1 and see both side equal , then that's proved
but when i try another n=2 then it s not equal
for n=1 we have (3*1-2)^2=1^2=1 on lhs we have 1*(6*1^2-3*1-1)/2=2/2=1 on rhs so now we get to assume is true for k>=1 and now we have to show it is true for k+1
1^2+4^2+7^2+....+(3k-2)^2+(3(k+1)-2)^2 we want to show this equals the other side (whereever this is n we need k+1 to be there)
so we know 1^2+4^2+...+(3k-2)^2=k(6k^2-3k-1)/2 so we have k[6k^2-3k-1]/2+(3(k+1)-2)^2 we need to manipulate this so that it is (k+1)*(6(k+1)^2-3(k+1)-1)/2 ok? try to do this
oh hello, n=2 it does work for n=2 we have on lhs 1^2+4^2=1+16=17 on rhs we have 2(6*2^2-3*2-1)/2=24-6-1=17 so does work for n=2
it wont work for the k+1 because the LHS ^ 2 and the RHS ^ 3 ? So can you explain more about this ? :-<
what do you mean?
let me try this again
i added in alot of zeroes i forced it to be what i want and then the left overs canceled with the other left overs :)
do you have any questions?
i got it now really appreciate for your help
so you actually understood all the weird stuff i did? lol
no i did it another way and got the answer
if i looked back on this, i would be like wtf
lol i was trying to do it another way and i kept failing
can you post what you did
i dont know how to post the paper attaach in here , and i dont have the scan machine :(
but i will try to type it then
no worries lol