2 Math Induction Questions

Ahhh...math induction...WTH is this stuff. Anyone have any good links for easy ways to learn it?

Anyway...here are the problems:

For ecah natural number n, with n greater than or equal to 6, 2^{n} > (n+1)^{n} **I was thinking since P(k) is tue, then 2^{k}>(k+1)^{2} multiply both sides by two and we get: 2^{k}*2 > 2(k^{2}+2k+1) and I am stuck...

2nd question:

Prove or disprove:

For each natural number n, 1/(1*2)+1/(2*3)+...+1/n(n+1)=n/n+1

Any help would be great!!

Re: 2 Math Induction Questions

1.) Prove by induction that:

$\displaystyle 2^n>(n+1)^2$ where $\displaystyle 6\le n\in\mathbb{N}$

First step, show that the base case $\displaystyle P_6$ is true:

$\displaystyle 2^6>(6+1)^2$

$\displaystyle 64>49$ true.

State the induction hypothesis $\displaystyle P_k$:

$\displaystyle 2^k>(k+1)^2$

Multiply through by 2:

$\displaystyle 2^{k+1}>2(k+1)^2$

Now, can you show that:

$\displaystyle 2(k+1)^2>((k+1)+1)^2$ ?

2.) Prove by induction that:

$\displaystyle \sum_{j=1}^n\frac{1}{j(j+1)}=\frac{n}{n+1}$ where $\displaystyle n\in\mathbb{N}$.

Show that the base case $\displaystyle P_1$ is true.

State the induction hypothesis $\displaystyle P_k$:

$\displaystyle \sum_{j=1}^k\frac{1}{j(j+1)}=\frac{k}{k+1}$

Now, what should you add to both sides?

Re: 2 Math Induction Questions

Quote:

Originally Posted by

**Measure13** Ahhh...math induction...WTH is this stuff. Anyone have any good links for easy ways to learn it?

Anyway...here are the problems:

For ecah natural number n, with n greater than or equal to 6, 2^{n} > (n+1)^{n} **I was thinking since P(k) is tue, then 2^{k}>(k+1)^{2} multiply both sides by two and we get: 2^{k}*2 > 2(k^{2}+2k+1) and I am stuck...

2nd question:

Prove or disprove:

For each natural number n, 1/(1*2)+1/(2*3)+...+1/n(n+1)=n/n+1

Any help would be great!!

Think of a row of dominos. You know that any domino will push the next one over, as long as the first one is pushed.

Mathematical induction works the same way. If you want to prove that a statement is true for all possible values, then you first need to "push the first domino" by showing the statement is true for the first value. You then need to show that "pushing any domino will push the next one over", which means that you assume the statement is true for an arbitrary value and use this to show that the statement is true for the next value.

Re: 2 Math Induction Questions

Quote:

Originally Posted by

**MarkFL2** 1.) Prove by induction that:

$\displaystyle 2^n>(n+1)^2$ where $\displaystyle 6\le n\in\mathbb{N}$

First step, show that the base case $\displaystyle P_6$ is true:

$\displaystyle 2^6>(6+1)^2$

$\displaystyle 64>49$ true.

State the induction hypothesis $\displaystyle P_k$:

$\displaystyle 2^k>(k+1)^2$

Multiply through by 2:

$\displaystyle 2^{k+1}>2(k+1)^2$

Now, can you show that:

$\displaystyle 2(k+1)^2>((k+1)+1)^2$ ?

2.) Prove by induction that:

$\displaystyle \sum_{j=1}^n\frac{1}{j(j+1)}=\frac{n}{n+1}$ where $\displaystyle n\in\mathbb{N}$.

Show that the base case $\displaystyle P_1$ is true.

State the induction hypothesis $\displaystyle P_k$:

$\displaystyle \sum_{j=1}^k\frac{1}{j(j+1)}=\frac{k}{k+1}$

Now, what should you add to both sides?

Thanks boss!