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?

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 $\displaystyle \sum\limits_{n = 1}^N {\left( {2n - 1} \right)} = N^2 $.

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

The separate inductions should be easier.