Scripting Help!

[KoubaK]

hey, guys im new to UNIX and i need help.. I need to write a script which will tell you if you entered an even (e.g., 2, 4, 6...) or odd number (e.g., 1, 3, 5...) and then give you the summation of the number you entered. For example, if you entered 5 it will give 15 (1+2+3+4+5). I'm suppose to use if and loops .. I got no idea on how to do so.. below is what I got sooo far!

#!/bin/bash

echo " This script will tell you if you entered an even (e.g., 2, 4, 6...) or odd number (e.g., 1, 3, 5...) and then give you the summation of the number you entered. For example, if you entered 5 it will give 15 (1+2+3+4+5)"

echo "Please, enter your name and PRESS ENTER"
read name

echo 'Please, enter an integer number (e.g., 1, 2, 3...) and PRESS ENTER:'
fi

and I am to follow this example!

This script will tell you if you entered an even (e.g., 2, 4, 6...) or odd number (e.g., 1, 3, 5...) and then give you the summation of the number you entered. For example, if you entered 5 it will give 15 (1+2+3+4+5)

Please, enter your name and PRESS ENTER
name

Please, enter an integer number (e.g., 1, 2, 3...) and PRESS ENTER: 6
         name you have entered 6 please see the results

         The number you entered 6 is an even number
         The summation of 6 is 21

anyone? please help!! lol

[lmo]

#!/bin/bash
read -p "Please enter your name: " name
read -p "$name Please, enter an integer number (e.g., 1, 2, 3...) and PRESS ENTER: " number
digs=$number
sum=0
while [ $digs -gt 0 ];do sum=$(( sum + $digs )); digs=$(( $digs -1 )); done
echo "The summation of $number is $sum and I forgot about even numbers"
if [ $(( $number % 2 )) -eq 0 ];then
  echo "$number is an even number"
else
  echo "$number is not an even number"
fi
echo " see: 'man bash'  and 'man test' for an enormous amout of information."

[KoubaK]

thanks man!!! and thanks for the extra comment!! :P guess I will have too!!