SOLUTION: Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. 1. 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = 4(

4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = 4(4n+1)(8n+7)/6

is false because it isn't even true when n=1.

4 ⋅ 6  = 4(4*1+1)(8*1+7)/6

   24 = 4(5)(15)/6

   24 = 50

Also 4n(4n+2) is not even the correct formula
for the nth term.  The correct nth term formula
of 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ...
is (n+3)(n+5).  So it's wrong all the way around.

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 1²+4²+7²+∙∙∙+(3n-2)² = n(6n²-3n-1)/2

That appears to be correct.
 
If n=1 then  1² = 1(6∙1²-3∙1-1)/2 = 1(6-3-1)/2 = 1(2)/2 = 1 which 
shows that the proposition is true when n=1.

Let us assume that n=k is some integer (perhaps 1) such that
1²+4²+7²+∙∙∙+(3k-2)² = k(6k²-3k-1)/2.  

We want to show that the expression n(6n²-3n-1)/2 with k+1 substituted
for n also holds for the sum of the first k+1 terms.  That is, we want 
to be able to erase the question mark here: 

 1²+4²+7²+∙∙∙+[3(k+1)-2]² ≟ (k+1)[6(k+1)²-3(k+1)-1]/2 

or upon multiplying out the right side (you do that)

 1²+4²+7²+∙∙∙+[3(k+1)-2]² ≟ 3k³+15k²/2+11k/2+1

We start with our assumption:

1²+4²+7²+∙∙∙+(3k-2)² = k(6k²-3k-1)/2.

or

1²+4²+7²+∙∙∙+(3k-2)² = 3k³-3k²/2-k/2

and add [3(k+1)-2]² to both sides:

1²+4²+7²+∙∙∙+(3k-2)²+[3(k+1)-2]² = k(6k²-3k-1)/2 + [3(k+1)-2]²

Multiplying the right side all the way out, (you do that) 
we get 

1²+4²+7²+∙∙∙+(3k-2)²+[3(k+1)-2]² = 3k³+(15/2)k²+(11/2)k+1

and that is the same right side as the right side
of above when we had the equal sign with the question
mark above it.

1²+4²+7²+∙∙∙+[3(k+1)-2]² ≟ 3k³+(15/2)k²+(11/2)k+1 

Edwin