Question

prove the statement true :
1^2 +4^2 + 7^2 + ..... + (3n-2)^2 = [n (6n^2 - 3n -1)] / 2

Answer 1

did you see my earlier reply to this?

Answer 2

i dont see it :(

Answer 3

oh wait nvm ignore the early one do you know anything about induction?

Answer 4

some1 tell me put n=1 and see both side equal , then that's proved

Answer 5

but when i try another n=2 then it s not equal

Answer 6

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

Answer 7

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)

Answer 8

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

Answer 9

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

Answer 10

it wont work for the k+1 because the LHS ^ 2 and the RHS ^ 3 ? So can you explain more about this ? :-<

Answer 11

?

Answer 12

what do you mean?

Answer 13

let me try this again

Answer 14

here i will scan what i did.
you really have to know some algbra for this

Answer 15

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 :)

Answer 16

do you have any questions?

Answer 17

i got it now
really appreciate for your help

Answer 18

so you actually understood all the weird stuff i did? lol

Answer 19

no i did it another way and got the answer

Answer 20

if i looked back on this, i would be like wtf

Answer 21

LOLLLLL

Answer 22

lol i was trying to do it another way and i kept failing

Answer 23

can you post what you did

Answer 24

i dont know how to post the paper attaach in here , and i dont have the scan machine :(

Answer 25

but i will try to type it then

Answer 26

no worries
lol