Use? Gauss's approach to find the following sums?/ math help

Question

Use? Gauss's approach to find the following sums? (do not use? formulas).

 

a. 1+2+3+4+...99

b. 1+3+5+7+...1001

 

a. The sum of the sequence is?

b. ?

Answer 1

a. 1+2+3+4+...99

a. The sum of the sequence is?

 

This problem is: "Add the first n integers"

Solution Method (à lá Gauss):  Let's add them twice:

    1  +   2   +   3   +   4   +   5

    5  +   4   +   3   +   2   +   1

-------------------------------------

    6  +   6   +   6   +   6   +   6      There are five 6's (that is,

                                               there are n (n+1)'s

                          But we added them twice, so take half the total.

                            (Why bother with even/odd?)

 

This process is modeled by the math formula:   (n)(n+1)/2

 

However, if we were given a table of sums and were told to find Sum(5) and it was 15.  That is a different process.

    n       Sum

    1         1

    2         3

    3         6

    4        10

    5        15

    6        21

    7        28

  ...

 

A "table look-op" is a different process.  It assumes that the calculation has been done (or an estimate made) previously.  This how trig functions are done in many computer chips -- the manufacturer includes a read-only memory table with lots of values already calculated (and the story of the wrong value being stored in early manufactured Pentium chips makes sense).

 

An example:  The SORTing process has numerous methods (processes) but no single formula, however it is clearly easier to re-arrange the names on a sheet of paper than to get students to "line up in alphabetical order."

 

So, the problem "1+2+3+4+...+99" produces ninety-nine 100's, but we added the sum twice, so now take half.

 

b. 1+3+5+7+...1001
b. ?

 

From Part A (using old knowledge is a great process !), we have 1001  1002's, so we take half the sum.  Now, we subtract the even values; but, how many are there?  The missing numbers are 2+4+6+8+...+1000, which is exactly:

      2*(1+2+3+4+...+500).  We may use even/odd information here.

 

Again, multiplying 1002 by 1000, then dividing by 2, then, multiplying 501 by 500 and dividing&multiplying by 2, then subtracting the sums is both a process and a formula.

Answer 2

Well, Gauss approach has be "formalized."

a) (99/2)(1 + 99)

b) (1001/2)(1 + 1001) / 2           divide by 2 because only the odds are given

Answer 3

1.  Gauss (as a youngster) would have paired the first and last numbers of the first series to make 1 + 99 = 100, and the second and next-to-last to make 2+98 = another 100, and then would have had 49 hundreds = 4900 plus one fifty in the middle not yet counted, so he'd get 4950 as the sum. He would have done this mentally.

 

2. Try a similar approach, 'pairing from L & R', on the second problem.

 

Formulas for sums of arithmetic series have been ruled out in the instructions!