SOLUTION: Please help me solve and explain how to solve this mathematical induction: Use mathematical induction to prove that the following is true for every positive integer n: 1/4+2*

This problem is really botched up.  You should point it out to your
teacher.  This is what you have above:

But your 2nd and 3rd terms are not what we get when we substitute
2 and 3 in the given nth term [2(n-1)+1/4]

Substituting n=1 in [2(n-1)+1/4] = [2(1-1)+1/4] = [2(0)+1/4] = 1/4,
That's OK for you do have 1/4 for the 1st term.  However:

Substituting n=2 in [2(n-1)+1/4] = [2(2-1)+1/4] = [2(1)+1/4] = 9/4
That's not what you have for the 2nd term, 2*(1/4), for that's 1/2

Substituting n=3 in [2(n-1)+1/4] = [2(3-1)+1/4] = [2(2)+1/4] = 17/4
That's not what you have for the 3rd term, 4*(1/4), for that's 1.

Also the nth term on the left is not simplified:

So let's simplify your given nth term [2(n-1)+1/4]:

2(n-1)+1/4
2n-2+1/4
8n/4-8/4+1/4
(8n-8+1)/4
(8n-7)/4

So the left side of your sequence should be like the left side below

1/4 + 9/4 + 17/4 + ... + (8n-7)/4 = (4n²-3n)/4

The right side is OK.  But since there is a 4 in the denominator of every term
on the left and also on the right side, we can just prove that 4 times the left
side equals 4 times the right side.  Then after we've proved that we'll just
divide both sides by 4.

So all we need do is prove this by induction:

1 + 9 + 17 + ... + (8n-7) = 4n²-3n

and then divide everything through by 4.

Induction works like this:

If a formula is assumed true for one value of n, and that assumption
leads to its truth for the NEXT velaue of n, then all we have to do
is prove that it is true when n=1, and then it will be true for n=2.
And then the fact that it is true for n=2 will show that it is true
for n=3, and so on, for all integers.

1 + 9 + 17 + ... + (8n-7) = 4n²-3n

We show that it is true when n=1

When n=1 we have just the first term 1 on the left.

On the right we have 4(1)²-3(1) = 4-3 = 1

So the formula is true when n=1

So therefore we know it is true for at least one value of n.

So we want to show that if it is assumed true for one value of n, say n=k, then
it is true for the next value of n=k+1 as well.  So we assume it is true for 
some value of n, n=k.  So we assume this is true:

1 + 9 + 17 + ... + (8k-7) = 4k²-3k

What we need to prove now is that when we add the next term 
[the (k+1)st term] to both sides, like this:

1 + 9 + 17 + ... + (8k-7) + [8(k+1)-7] = 4k²-3k + [8(k+1)-7]

that we will get the same thing as if we substituted k+1 for n in the
formula:  1 + 9 + 17 + ... + (8n-7) = 4n²-3n

which would be

1 + 9 + 17 + ... + [8(k+1)-7] = 4(k+1)²-3(k+1)

So essentially we want to prove that adding the (k+1)st term to the formula
gives the same expression as substituting (k+1) into the formula. So we
want to show that 

4k²-3k + [8(k+1)-7] is equal to 4(k+1)²-3(k+1)

If we can work them both down to the same thing, then we will have proved the
problem by induction:

4k² - 3k + [8(k+1)-7]  ----------  4(k+1)²-3(k+1)    

4k² - 3k + [8k+8-7]    ----------  4(k²+2k+1)-3k-3

4k²-3k+8k+1            ----------  4k²+8k+4-3k-3
    
4k²+5k+1               ----------  4k²+5k+1  

So they both worked out to be the same. so we have proved this:

1 + 9 + 17 + ... + (8n-7) = 4n²-3n

So all we have to do to prove your original problem,
which was botched by the the 2nd and 3rd terms being
wrong and the nth term not simplified, is to divide
both sides of what we proved, by 4:

1/4 + 9/4 + 17/4 + ... + (8n-7)/4 = (4n²-3n)/4

I cannot imagine why you were given such a messed-up problem
to prove by induction.  Please point it out to your teacher.

Edwin