Hi I am having alot of trouble with this induction question. Any help would be much appreciated!
Find a formula for 1/1*2 + 1/2*3 + 1/n*(n+1) and prove by induction.
Here's some details. You'll notice that each term can be split up
$\displaystyle \frac{1}{1 \cdot 2} = \frac{1}{1} - \frac{1}{2},\;\;\;\;\;\frac{1}{2 \cdot 3} = \frac{1}{2} - \frac{1}{3},\;\;\;\;\;\frac{1}{3 \cdot 4} = \frac{1}{3} - \frac{1}{4} $
$\displaystyle \frac{1}{n \cdot (n+1)} = \frac{1}{n} - \frac{1}{n+1} $
Then when you add them up all the terms cancel except the first and last, so
$\displaystyle S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}$
For the induction part, it true for the first few, assume true for $\displaystyle n = k$
so
$\displaystyle S_k = \frac{k}{k+1}$
Prove true for $\displaystyle n = k+1$, i.e. $\displaystyle S_{k+1} = \frac{k+1}{k+2}$
so
$\displaystyle S_{k+1} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} $
$\displaystyle S_{k+1} = \frac{k+1}{k+2} $
so it must be true for all n.
Could you verify that I am on the right track for another?
Prove that 3^n < n! if n is an integer greater than 6..
- Base case 3^7 = 2187 and 7! = 5040
- Inductive
- Assume for p(k) show for p(k+1)
- 3^K+1 < K+1!
- 3^k + 3^k < K + 1!
I am stuck on this part now.
You should have an initial condition that n>2
$\displaystyle
3^{n} < n!
$
Let the following is true
$\displaystyle P(k) = 3^{k}<k!$
To prove
$\displaystyle
P(k+1) = 3^{k+1}<(k+1)!
$
Therefore
$\displaystyle
3^{k}\times 3 < k!(k+1)
$
Putting k! instead of 3^k we get something greater hence if we can prove that this greater thing less than our RHS
then we have done it
$\displaystyle
3*k! < k! (k+1)
$
For $\displaystyle k > 2$
Thus $\displaystyle 3<(k+1)!$
and $\displaystyle
3k!<(k+1)!
$