Please explain this: 1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2
do you know these ? \(\sum n = \dfrac{n(n+1)}{2} \\ \sum n^2 = \dfrac{n(n+1)(2n+1)}{6}\) ?
Yes I did that and n= 1
\(\sum c=nc\)
n= 1 ?
To answer the prob it first said I had to prove it equals n=1
oh, the mathematical induction way!?
Yeah , sorry
i was doing it without induction....but oh well... so, for n=1, you got that true, right ?
Yes
so we assume it true for n= k 1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = k(6k^2-3k-1)/2
and prove it for n=k+1
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
Wat I don't get where u get (3k+1)^2 and (k+1)
we need to prove it for n = k+1 right ? so, 3 (k+1) -2 = 3k+3-2=3k +1 ok?
Ok then y are there two of them Plus and minus
sorry, what exactly is your doubt ?
Y is there both 3k+1 and 3k-1 . I can see how we got the plus not minus
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
(3k-1)^2
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 ?
I don't understand
i just replaced every 'k' by 'k+1' thats it!
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 ?
i still dont get how you get (3k-1)^2
but but i didn't get (3k-1)^2 at all i had (3k-2)^2 term though...
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
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 ?
thanks i think i get it
sorry if i confused you...
nah im good
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