anonymous
  • anonymous
prove the statement true : 1^2 +4^2 + 7^2 + ..... + (3n-2)^2 = [n (6n^2 - 3n -1)] / 2
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • myininaya
did you see my earlier reply to this?
anonymous
  • anonymous
i dont see it :(
myininaya
  • myininaya
oh wait nvm ignore the early one do you know anything about induction?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
some1 tell me put n=1 and see both side equal , then that's proved
anonymous
  • anonymous
but when i try another n=2 then it s not equal
myininaya
  • myininaya
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
myininaya
  • myininaya
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)
myininaya
  • myininaya
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
myininaya
  • myininaya
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
anonymous
  • anonymous
it wont work for the k+1 because the LHS ^ 2 and the RHS ^ 3 ? So can you explain more about this ? :-<
myininaya
  • myininaya
?
myininaya
  • myininaya
what do you mean?
anonymous
  • anonymous
let me try this again
myininaya
  • myininaya
here i will scan what i did. you really have to know some algbra for this
1 Attachment
myininaya
  • myininaya
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 :)
myininaya
  • myininaya
do you have any questions?
anonymous
  • anonymous
i got it now really appreciate for your help
myininaya
  • myininaya
so you actually understood all the weird stuff i did? lol
anonymous
  • anonymous
no i did it another way and got the answer
myininaya
  • myininaya
if i looked back on this, i would be like wtf
anonymous
  • anonymous
LOLLLLL
myininaya
  • myininaya
lol i was trying to do it another way and i kept failing
myininaya
  • myininaya
can you post what you did
anonymous
  • anonymous
i dont know how to post the paper attaach in here , and i dont have the scan machine :(
anonymous
  • anonymous
but i will try to type it then
myininaya
  • myininaya
no worries lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.