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Find a formula for 1+4+7+...+(3n-2)for positive integars n, and then verify your formula by mathematical induction.

So far this is what I've done:

Claim: 1+4+7+...+(3n-2)=n^2+(1/2)(n^2-n) for every positive integar n.

Proof: Induction on n

Base: Let n=1. Then n^2+(1/2)(n^2-n)= 1^2+(1/2)(1^1-1) = 1.

Step: Assume that for some postivie integar k,

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

Now we show that P(k+1) is true.

Thus 1+4+7+...+(3k-2)+(3(k+1)-2)= (k+1)^2+(1/2)[(k+1)^2-(k+1)]

This is where I am stuck because I don't understand what to do next with my equation. If you could help me out and let me know of any other errors I have in this proof so far that would be awesome! Thanks!

What you need to do is show that this expression is equal to the expression when n=k+1.

You have written "Thus 1+4+7+...+(3k-2)+(3(k+1)-2)= (k+1)^2+(1/2)[(k+1)^2-(k+1)]"

How is that true? You haven't proved it, you've simply asserted it as if it were true!

You need to take the previous statement, "P(k)" you might call it, and from this you have to DEDUCE that "P(k+1)" is true:

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

The left hand side needs one extra term to be the right thing:

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

Our goal is to show the right hand side here:

k^2 + (1/2)(k^2-k) + (3k+1)

= (3/2)k^2 - (5/2)k + 1

is equal to to:

(k+1)^2+(1/2)( (k+1)^2-(k+1) )

= k^2 + 2k + 1 + (1/2)( k^2 + 2k + 1 - k - 1 )

= k^2 + 2k + 1 + (1/2)( k^2 + k )

= (3/2)k^2 + (5/2)k + 1

Now, combining these two arithmetic messes, we have shown that if you assume P(k) is true:

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

Then add 3k+1 to both sides:

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

Then rearrange the expression on the right (as above):

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

which is P(k+1).

This is proof that P(k)

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