Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2

Answer 1

using http://openstudy.com/study#/updates/504b9d7ce4b0985a7a58b494 I understand everything until

Answer 2

I don't get how the last form is "precisely what we were supposed to get" and how it proves it true

Answer 3

hmmm... first we need the basis
then when we have k
and then k+1
have to manipulate the left to equal the right ..

Answer 4

for n= 1, we have:
left side = 1^2 = 1
right side= 1*(6-3-1)/2= 1
so our proposition is true for n=1

Answer 5

ok so we have the basis proven true.. next we need to have n = k

Answer 6

in general, the mathematical induction principle requests another checking, for example n=2

Answer 7

Ok I understand why and how to do that

Answer 8

so, for n=2, we have:
left side = 1+ 4^2=1+16=17
right side = 2*(24-6-1)/2= 17

Answer 9

but what about n = k + 1?
how is the simplified form of n = k+1 proving it true???

Answer 10

first we need n = k and then we go to n = k+1 . That guy went forward to n+1

Answer 11

when we have a proposition which depends on a natural number, say n, namely, we have:
P(n), then if P(1) is true and P(n+1) is true when P(n) is true, then we can state, that P(n) is true for all natural numbers, namely:
\[\Large \forall n \in \mathbb{N}\]

Answer 12

\[1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \frac{n(6n^2-3n-1)}{2}\]

for n = k

\[\[1^2 + 4^2 + 7^2 + ... + (3k - 2)^2 = \frac{k(6k^2-3k-1)}{2}\]\]

for n = k+1

\[\[1^2 + 4^2 + 7^2 + ... + (3(k+1) - 2)^2 = \frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2}\]\]

Answer 13

so to simplify this mess a bit ... sorry this is long!

\[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6(k+1)^2-3(k+1)-1)}{2} \]

\[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6(k+1)^2-3k-4)}{2} \]

\[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6(k^2+2k+1)-3k-4)}{2} \]

\[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6k^2+12k+6)-3k-4)}{2} \]

\[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6k^2+12k-3k-4+6)}{2} \]
\[\[1^2 + 4^2 + 7^2 + ... + (3k+1)^2 = \frac{(k+1)(6k^2+9k+2)}{2} \]

Answer 14

hmmm then manipulate the left... hold on this is in my book I need to refresh my memory on it.. I remember substituting that 1^2+4^2+7^2

Answer 15

I know that our goal is to have the left = right ... I just have to see one of the examples in my book again.

Answer 16

a proposition was used...

For all n in N
then the entire 1^2 4^2 7^2 is substituted .

we do know that \[(3k+1)^2 = (3k+1)(3k+1) = 9k^2+6k+2 \]