SOLUTION: Use mathematical induction to prove the following: For each natural number n,1+5+9+....(4n-3)=n(2n-1).

1+5+9+...+(4n-3) = n(2n-1)

Note: The last term is (4n-3)

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It is only necessary to show the formula works
for n=1 before showing that if it works for n = k
it works for n = k+1  but I will show it works for
1, 2 and 3.

It works for n=1 because

Last term = 4n-3 = 4*1-3 = 4-3 = 1

1 = 1(2*1-1) 
1 = 1(2-1)
1 = 1(1)
1 = 1

It works for n=2 because

Last term = 4n-3 = 4*2-3 = 8-3 = 5

1+5 = 2(2*2-1) 
1+5 = 2(4-1)
1+5 = 2(3)
1+5 = 6

It works for n=3 because

Last term = 4n-3 = 4*3-3 = 12-3 = 9

1+5+9 = 3(2*3-1) 
1+5+9 = 3(6-1)
1+5+9 = 3(5)
1+5+9 = 15

Now we assume that k is some integer for which
the formula works (and we have 3 such values 
already for which we know the formula works.)

So we assume

1+5+9+....(4k-3) = k(2k-1).

What we want to show is that if we substitute k+1 for n in

1+5+9+...(4n-3) = n(2n-1)

It will also work.

That is we want to show that based on the assumption that it 
works for k, that we'll end up with this without the question
mark over the equal sign:

1+5+9+...+[4(k+1)-3] ≟ (k+1)[2(k+1)-1]
                
The question mark above the equal sign is to show that we
have not shown that yet.  The above is what we must show.

We will simplify what we are to show:

1+5+9+...+[4k+4-3] ≟ (k+1)[2k+2-1]

1+5+9+...+(4k+1) ≟ (k+1)(2k+1)

1+5+9+...+(4k+1) ≟ 2k²+3k+1

Now we will show it.  We start with:

1+5+9+...+(4k-3) = k(2k-1)

We add the k+1st term, which is (4k+1), to both sides,
and hope that the right side comes out to to the
expression above, which is 2k²+3k+1:

1+5+9+...+(4k-3)+(4k+1) = k(2k-1)+(4k+1)
1+5+9+...+(4k-3)+(4k+1) = 2k²-k+4k+1 
1+5+9+...+(4k-3)+(4k+1) = 2k²+3k+1

So we have shown that if the formula works for some
value of n, say n=k, then it will always work for the
next integer n=k+1.

Therefore
Since we have shown it works for n=k=1, it works for n=k+1=2
Since it works for n=k+1=2, it works for n=k+2=3
Since it works for n=k+2=3, it works for n=k+3=4
Since it works for n=k+3=4, it works for n=k+4=5
etc. etc. forever

Edwin