1. mathematical induction

use mathematical induction to prove the formula for every positive integer n.

3 + 5 + 7 + ... + (2n +1)= n(n +2)

this is a beastly problem.

i have proven that

n=1 so 2(1)+1=1(1)+2
3=3 so this works but now im dont know what to do next. i know im suppose to go further but im so lost. please help me understand this.

2. Skip the trivial case and assume true for

$1+3+5+......+(2k+1)=k(k+1)$

Show $P_{k+1}$ is true:

$1+3+5+....+(2k+1)+(2k+3)=k(k+2)+2k+3=(k+1)(k+3)$

Therefore, it is true for $P_{k+1}$

3. Originally Posted by flexus
use mathematical induction to prove the formula for every positive integer n.

3 + 5 + 7 + ... + (2n +1)= n(n +2)

this is a beastly problem.

i have proven that

n=1 so 2(1)+1=1(1)+2
3=3 so this works but now im dont know what to do next. i know im suppose to go further but im so lost. please help me understand this.
Assume true for n, now you have to prove it for n+1, i.e. that:
$3+5+7+...+(2n+1) + (2(n+1)+1) = (n+1)(n+3)$

And you'd do that by saying:

$3+5+7+...+(2n+1) + (2(n+1)+1) = n(n+2) + (2(n+1)+1)$

and working from there.

Because you have *assumed* that it's true for n, you can directly substitute the $n(n+2)$ term for the series up to $(2n+1)$.

And having *assumed* that it's true for n, (that bit's called the "induction hypothesis"), you will have *proved* that it still holds for n+1 (that bit's called the "induction step" and it's usually the one where the algebra happens).

So once you've shown it true for the base case, and by assuming the truth of the induction hypothesis for n you've shown it still holds for n+1, "the result follows by the Principle of Mathematical Induction".

4. Originally Posted by galactus
Skip the trivial case and assume true for

$1+3+5+......+(2k+1)=k(k+1)$

Show $P_{k+1}$ is true:

$1+3+5+....+(2k+1)+(2k+3)=k(k+1)+2k+3=(k+1)(k+3)$

Therefore, it is true for $P_{k+1}$
In your second line of maths I think you mean $k(k+2)$ not $k(k+1)$.

5. Originally Posted by Matt Westwood
In your second line of maths I think you mean $k(k+2)$ not $k(k+1)$.
Yes, that was a typo. Thanks for the heads up.

6. Originally Posted by galactus
Skip the trivial case and assume true for

$1+3+5+......+(2k+1)=k(k+1)$

Show $P_{k+1}$ is true:

$1+3+5+....+(2k+1)+(2k+3)=k(k+2)+2k+3=(k+1)(k+3)$

Therefore, it is true for $P_{k+1}$
lol man your avatar is funny but also crazy haha