Question

I got the answer solved for S1, but am having problems with S2 adn S3, making them true. I am following all the steps but it just won't work!
Sn: 1^2+4^2+7^2+...+(3n-2)^2 = n(6n^2-3n-1)/2.

Answer 1

what did you get for S1 ?

Answer 2

1=1

Answer 3

I just plugged in 1^2

Answer 4

you should "plug in" 1 (not 1^2) (though 1^2 = 1 , so it still works)

Answer 5

Let me try that first. Hold on.

Answer 6

S2 is short for use n=2 in the formula
n(6n^2-3n-1)/2

Answer 7

Okay for S2, I got 1,1) so it is true.2

Answer 8

So for S3, I assume I plug in 2?

Answer 9

S1 means use n=1 (and you will get 1 from the formula)
for S2, use n=2
what do you get ?

Answer 10

Hold on. For S1, I got 1, 1. Let me try 2.

Answer 11

using 2 in S2, I got 16=18. Not true.

Answer 12

Sorry, I meant 16=17

Answer 13

the "left side" goes from
1^2 upto (3n-2)^2
when n=1 the last term is (3*1-2)^2 = (3-2)^2 = 1^2
so the series for S1 is just
S1= 1^2


for S2, the last term is (3*2 - 2)^2 = (6-2)^2 = 4^2
and the series is
S2= 1^2 + 4^2

if we simplify that we get 1+16= 17
so S2 = 17

now use the formula (the right side)
replace n with 2 and get
\[ \frac{2(6\cdot 2^2-3\cdot 2-1)}{2} \]

Answer 14

Having battery issues with calculator. Hold on.

Answer 15

16=16

Answer 16

no, you should get 17

remember, the series is a sum of n terms
in other words, when n=2 , S2 is the sum of 2 terms:
1^2 + 4^2
which adds up to 17 (1+16)

the formula, if you are careful, also gives 17

Answer 17

do it step by step. for n=2 the formula (after replacing all the n's with 2, becomes
\[ \frac{2(6\cdot 2^2-3\cdot 2-1)}{2} \]
inside the parens we have
6*2*2 - 6 - 1
or
24 -7
or
17

so you have
\[ \frac{2(17)}{2} \]

Answer 18

Okay. that is interesting.

Answer 19

I just redid it and got 17=17. Thanks.

Answer 20

Thanks.

Answer 21

can you do S3? the left side will have 3 terms
1^2+ 4^2 + 7^2

Answer 22

fyi, when they write
Sn: 1^2+4^2+7^2+...+(3n-2)^2

the (3n-2)^2 is the *last term*
when n is 3, we get the last term= (3*3-2)^2 = (9-2)^2 = 7^2
that means S3 is
S3= 1^2+4^2+7^2

also, S4= 1^2+4^2+7^2+10^2,
and S5= 1^2+4^2+7^2+10^2 + 13^2
and so on.

Answer 23

Thank you. I just did S3 and got 67=67.

Answer 24

I get 66

Answer 25

the left side is
S3= 1^2+4^2+7^2
= 1 + 16 + 49
= 50+16
= 66

Answer 26

Let me retry. I think I must of miscalculated a number someplace.

Answer 27

it helps to write it down, and simplify in small steps.

Answer 28

I still get 67=67. I will continue trying. Thank you phi. I do appreciate your help. But I will keep trying. Let me take a break and retry it again. Sometimes you need to walk away for a little bit. Been at 3 problems for 3 hours and this is the one wiht the problem.

Thanks.

Answer 29

how do you get 67 from the left side which is
1^2+4^2+7^2
?