# Proof by Mathematical Induction

• Jul 24th 2011, 01:55 PM
lovesmath
Proof by Mathematical Induction
Prove that for all integers n >= 1,

1/3 = (1 + 3)/(5 + 7) = (1 + 3 + 5)/(7 + 9 + 11) = ...
= (1 + 3 + ... + (2n - 1))/((2n + 1) + ... + (4n - 1))

This is what I have so far:
Show that P(1) is true. Left-hand side = 1/3. Right-hand side = 1/3. Since the left-hand side is equal to the right-hand side, P(1) is true. Show that for any integer k >= 1, if P(k) is true, then P(k + 1) is also true.
Suppose that k is any integer with k >= 1.

I am confused about how to complete the inductive step. Can anyone help me with this?
• Jul 24th 2011, 02:33 PM
Plato
Re: Proof by Mathematical Induction
Quote:

Originally Posted by lovesmath
Prove that for all integers n >= 1,
1/3 = (1 + 3)/(5 + 7) = (1 + 3 + 5)/(7 + 9 + 11) = ...
= (1 + 3 + ... + (2n - 1))/((2n + 1) + ... + (4n - 1))

I would do it in two parts.
The numerator $\sum\limits_{n = 1}^N {\left( {2n - 1} \right)} = N^2$.

The denominator $\sum\limits_{n = N}^{2N-1} {\left( {2n + 1} \right)} = 3 N^2$

The separate inductions should be easier.