1. ## induction problem

I have to prove that for all integers n>= 1

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

don't know how to go about it.

2. Hello, NidhiS

The sum of the first $\displaystyle n$ terms of an arithmetic series is given by:

. . $\displaystyle S_n \;=\;\frac{n}{2}[2a + (n-1)d]$ . . . where: .$\displaystyle \begin{Bmatrix}n &=& \text{no. of terms} \\ a &=& \text{first term} \\ d&=&\text{common diff.} \end{Bmatrix}$

Prove for all integers $\displaystyle n \geq 1\!:\;\;$

$\displaystyle \frac{1}{3} \;=\;\frac{1+3}{5+7} \;=\; \frac{1+3+5}{7+9+11}) \;=\; \hdots \;=\;\frac{1+3+\hdots + (2n-1)}{(2n+1)+(2n+3)+\hdots+(4n-1)}$

The numerator is: .$\displaystyle 1 + 3 + 5 + \hdots + (2k-1)$

This is an arithmetic series with: .$\displaystyle a = 1,\;d = 2,\;n = k$

. . Its sum is: .$\displaystyle N \;=\;\frac{k}{2}[2(1) + (k-1)2] \;=\;k^2$

The denominator is: .$\displaystyle (2k+1)+(2k+3) + \hdots + (4k-1)$

This is an arithmetic series with: .$\displaystyle a = (2k+1),\;d = 2,\;n = k$

. . Its sum is: .$\displaystyle D \;=\;\frac{k}{2}[2(2k+1) +(k-1)2] \;=\;3k^2$

Therefore: .$\displaystyle \frac{N}{D} \;=\;\frac{k^2}{3k^2} \;=\;\frac{1}{3}$