Ask your own question, for FREE!
Mathematics 10 Online
OpenStudy (anonymous):

Please explain this: 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2

hartnn (hartnn):

do you know these ? \(\sum n = \dfrac{n(n+1)}{2} \\ \sum n^2 = \dfrac{n(n+1)(2n+1)}{6}\) ?

OpenStudy (anonymous):

Yes I did that and n= 1

hartnn (hartnn):

\(\sum c=nc\)

hartnn (hartnn):

n= 1 ?

OpenStudy (anonymous):

To answer the prob it first said I had to prove it equals n=1

hartnn (hartnn):

oh, the mathematical induction way!?

OpenStudy (anonymous):

Yeah , sorry

hartnn (hartnn):

i was doing it without induction....but oh well... so, for n=1, you got that true, right ?

OpenStudy (anonymous):

Yes

hartnn (hartnn):

so we assume it true for n= k 1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = k(6k^2-3k-1)/2

hartnn (hartnn):

and prove it for n=k+1

hartnn (hartnn):

so we need to prove 1^2 + 4^2 + 7^2 + ... + (3k - 2)^2+ (3k+1)^2 = (k+1)(6[k+1]^2-3(k+1)-1)/2

OpenStudy (anonymous):

Wat I don't get where u get (3k+1)^2 and (k+1)

hartnn (hartnn):

we need to prove it for n = k+1 right ? so, 3 (k+1) -2 = 3k+3-2=3k +1 ok?

OpenStudy (anonymous):

Ok then y are there two of them Plus and minus

hartnn (hartnn):

sorry, what exactly is your doubt ?

OpenStudy (anonymous):

Y is there both 3k+1 and 3k-1 . I can see how we got the plus not minus

hartnn (hartnn):

i don't see a 3k-1 here 1^2 + 4^2 + 7^2 + ... + (3k - 2)^2+ (3k+1)^2 = (k+1)(6[k+1]^2-3(k+1)-1)/2

OpenStudy (anonymous):

(3k-1)^2

hartnn (hartnn):

its (3k-2)^2 ! and it will come because we are considering k+1 terms instead of k terms so (3k-2)^2 for k'th term and (3k+1)^2 for 'k+1'th term ! got this ?

OpenStudy (anonymous):

I don't understand

hartnn (hartnn):

i just replaced every 'k' by 'k+1' thats it!

hartnn (hartnn):

you replace k by k+1 in 1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = k(6k^2-3k-1)/2 what u get ?

OpenStudy (anonymous):

i still dont get how you get (3k-1)^2

hartnn (hartnn):

but but i didn't get (3k-1)^2 at all i had (3k-2)^2 term though...

OpenStudy (anonymous):

sorry i think im just overthinking but if you put (3(k+1)-2)^2 it equals (3k+1)^2, so how do you get (3k-2)^2

hartnn (hartnn):

it will come because we are considering k+1 terms instead of k terms so (3k-2)^2 for k'th term and (3k+1)^2 for 'k+1'th term ! got this ?

OpenStudy (anonymous):

thanks i think i get it

hartnn (hartnn):

sorry if i confused you...

OpenStudy (anonymous):

nah im good

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!