Question

12 + 42 + 72 + ... + (3n - 2)2 =n(6n^2-3n-1)/2

Answer 1

@jdoe0001 Mathematical Induction

Answer 2

Would like an elaborate answer or help me through what I;am supposed to do please?

Answer 3

well i think the problem can't be proven as it fails the test when n = 1

if you substitute n = 1 into the right hand side you get

\[\frac{1(6 \times 1^2 - 3 \times 1 - 1)}{2}= 1\]

so this means the sum of 1 term is 1

and the left hand side seems to say, the 1st term is 12...

which doesn't seem to work...

so it fails the 1st test for induction where n = 1

So can you please check that the information is correctly typed

Answer 4

I got 1

Answer 5

but it's actually 1^2+4^2 +7^2

Answer 6

I just paste it and didn't look to see if it is correct sorry

Answer 7

@campbell_st

Answer 8

@Zale101

Answer 9

try to add (3(n+1) - 2)^2 to the left side

Answer 10

n(6n^2-3n-1)/2 + (3(n+1) - 2)^2

Answer 11

see if you get (n+1)( 6(n+1)^2 - 3(n+1) - 1)/2

Answer 12

yes i did

Answer 13

I have (3(k+1)-2^2=k+1(6(k+1)^2-3(k+1)-1/2

Answer 14

@sourwing

Answer 15

n(6n^2-3n-1)/2 + (3(n+1) - 2)^2 = (1/2) (n+1) (6 n^2+9 n+2)
and
(n+1)( 6(n+1)^2 - 3(n+1) - 1)/2 = (1/2) (n+1) (6 n^2+9 n+2)

Answer 16

both sides are the same so, the formula is true

Answer 17

How did you do that?

Answer 18

wolfram alpha XD. It's just simple algebra so why do it by hand

Answer 19

ok... so if its 1^2 then you have proved it true to n = 1

now assume its true for n = k

so find the sum of k terms

Use the general form

\[\frac{n(6n^2 - 3n - 1)}{2}\]

so assume the kth term is

\[S_{k} = \frac{k(6k^2 - 3k - 1)}{2}\]

the last step to proving by induction is to show

\[S_{k + 1} =S_{k} + a_{k + 1}\]

so the k + 1 term is

\[(3(k + 1) -2)^2\]

so you job is to prove that right hand side is equal to the left hand side...


\[\frac{(k +1)(6(k + 1)^2 - 3(k + 1) -1)}{2} = \frac{k(6k^2 -3k -1)}{2} + (3(k + 1) -2)^2\]

hope this helps...