Problema Solution

Make a conjecture for each scenario.

1.) The sum of the first 100 positive odd numbers.

2.) The sum of two odd numbers.

3.) The product of two even numbers.

Answer provided by our tutors

1)The sum of the first 100 positive odd numbers.

 

 

1st odd number: 1

2nd odd number: 3

3rd odd number: 5 = 6 - 1

4th odd number: 7 = 8 - 1


The pattern is:

nth odd number: [2*n - 1]


Sum(nth odd number)=

Sum(2*n - 1)=

2 * Sum(n) - Sum(1)=

2 * Sum(n) - n


Moreover, we have:

Sum(n)= n * (n+1)/2


If n is even, this can be shown if you group the numbers this way:

(n+1) + ([n-1]+2) + ([n-2]+3) +...

You get n/2 groups of value (n+1).


Sum(nth odd number)=

2 * n * (n+1)/2 - n = n(n+1) -n = n*n


For n=100, the sum is:

100 * 100 = 10,000


---- ADDED ----

Here a simple trick to find the solution:

"The first 100 odd numbers form a series: 1,3,5,7,9...191,193,195,197,199. To determine the sum of all of the numbers in this series, pair off the first and last number, etc., which yields (1 and 199), (3 and 197), (5 and 195)...(97 and 103) and (99 and 101). Notice that the sum of each pair is 200. Since there are 100 numbers, there are 50 pairs; therefore, the sum is 200 times 50, or 10,000."

2.) The sum of two odd numbers.

         

               3+5=8

3.) The product of two even numbers.

            2*4=8