1. May 16, 2005

### evaboo

"Prove each of the following using Mathematical induction;" show all steps
pleasee someone help.. i have a test on this tommorow and i just need some examples.. could you also try to show all steps including the words so i understand how you got there? thakns so much in advance~!

1. -1/2, -1/4, -1/8... -1/2^n = (1/(2^n))-1

2. a + (a+d)+(a+2d)+...+[a+(n-1)d] = (n/2)[2a+(n-1)d]

3. 1^3 + 2^3 + 3^3... + n^3 = (n^2(n+1)^2)/(4)

4. show that (3^(4n))-1 is dividislbe by 80 for all positive integral values of n

2. May 16, 2005

### arildno

Let's take 3.
Part 1: Verification of the formula for n=1:
We have:
$$\frac{1^{2}(1+1)^{2}}{4}=\frac{1*4}{4}=1=1^{3}$$
That is, the formula is true for n=1.
Part 2: The induction step
Suppose it is true for n=k-1:
Then, we have:
$$1^{3}+2^{3}+++(k-1)^{3}+k^{3}=\frac{(k-1)^{2}k^{2}}{4}+k^{3}$$
since the formula holds for n=k-1.
Furthermore, we have:
$$\frac{(k-1)^{2}k^{2}}{4}+k^{3}=\frac{(k-1)^{2}k^{2}+4k^{3}}{4}=\frac{k^{2}((k-1)^{2}+4k)}{4}=\frac{k^{2}(k^{2}+2k+1)}{4}=\frac{k^{2}(k+1)^{2}}{4}$$
But that is precisely what the formula would predict it to be!

Thus, the formula is correct.

3. May 16, 2005

### evaboo

thanks!!! okay i did the same thing for number one and i got

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

how do i cancel this out?

4. May 16, 2005

### whozum

Lowest common denominator then separate 2^(k+1)

5. May 16, 2005

### evaboo

like 1/2^k + 1/2^k * 1/2 = 1/2^k - 2k/2k - (1/2^k * 1/2)?
where do i go from there?