SOLUTION: show that 8+16+24+...+8n = 4n(n+1) for all natural numbers. use mathematical induction.

Algebra.Com
Question 248658: show that 8+16+24+...+8n = 4n(n+1) for all natural numbers. use mathematical induction.
Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!
There are two parts to mathematical induction:
The first part is usually easy. For your formula, the first term is 8. Is the right side equal to 8 when n = 1? You should be able to find easily that the answer to this is "yes".

The second part usually takes the most work. First let's look at the left side of the formula for n:
8++16+...+8n
and look at the left side for (n+1):
8++16+...+8n+8(n+1)
The difference between the two, as we can see, is an extra term: 8(n+1). So we can take the formula for n:
8++16+...+8n = 4n(n+1)
and change the left side into the formula for (n+1) by adding 8(n+1) to each side:
8++16+...+8n+8(n+1) = 4n(n+1)+8(n+1)

Now we see if we can take this right side and, without changing the left side, change it into the formula for (n+1). It will help if we look ahead at what our goal is. The right side of the formula for n is: 4n(n+1). The right side of the formula for (n+1) will be what we get when we replace each n by (n+1):
4(n+1)((n+1)+1)
So our task now is to try to transform the right side we have now
4n(n+1)+8(n+1)
into
4(n+1)((n+1)+1)
without changing the left side of the equation.

There are probably many ways to do this. I notice that (n+1) is a factor of both what we have, 4n(n+1)+8(n+1), and what we want, 4(n+1)((n+1)+1), so I will start by factoring out (n+1) from 4n(n+1)+8(n+1) giving us:
8++16+...+8n+8(n+1) = (n+1)(4n+8)

I also notice that 4 is a factor of our goal. So I will factor out a 4 out of the (4n+8):
8++16+...+8n+8(n+1) = 4(n+1)(n+2)

The last step should be a bit obvious. We have the factors of 4 and (n+1) of our goal. The third factor of what we have is (n+2) and we want it to be ((n+1)+1). Since 2 = 1+1 we can change the (n+2) into (n+1+1). And since addition is associative, we can group it any way we want. So (n+2) = (n+1+1) = ((n+1)+1). Now we have:
8++16+...+8n+8(n+1) = 4(n+1)((n+1)+1)

In summary, we took the formula for n:
8++16+...+8n = 4n(n+1)
and, using regular Algebra, we changed it into the formula for (n+1):
8++16+...+8n+8(n+1) = 4(n+1)((n+1)+1)
This, combined with the earlier part where we showed that the formula works for the first term is our proof by mathematical induction.

Mathematical induction is like setting up dominoes to fall down. The hardest part is setting up the dominoes so that if one falls then it will knock down the next one. (This is the changing the formula for n into the formula for (n+1) part.) Then the last detail is to knock down the first domino. After all, no dominoes fall at all unless it gets started! (This is the "prove it for the first term part.)
RELATED QUESTIONS

Show, by the Principle of Mathematical Induction: 8 + 16 + 24+…+8n = 4n(n+1) for... (answered by stanbon)
Show by the Principle of Mathematical Induction: 8+16+24+…+8n=4n(n+1)for n≥1 (answered by ikleyn)
Use mathematical induction to prove the statement is true for all positive integers n.... (answered by ikleyn)
Pls help USE MATHEMATICAL INDUCTION TO PROVE THAT (n+1)^n < 2n^2 for all natural... (answered by ikleyn)
use the principal mathematical induction to show that the statement is true for all... (answered by Gogonati)
Use mathematical induction to prove the statement is true for all positive integers n, or (answered by Edwin McCravy)
Use Mathematical Induction to show that the following statement is true for all natural... (answered by stanbon,math_helper,ikleyn)
Use mathematical induction to prove the following: For each natural number... (answered by Edwin McCravy,amalm06)
Use the principle of mathematical induction to show that the statement are true for all... (answered by ikleyn)