## Sum Of 81297 81495 81693 100899 Solution

To derive the last equality, note that all the terms cancel pairwise, except the first term within the first parentheses and the last term within the last parentheses. This is a commonly used and powerful trick for calculating certain kinds of sums.

Example B.6

The arithmetic mean x of n numbers x\, x2, — xn is the sum of all the numbers divided by the number of terms, n:

Prove that n n n

Solution The difference x, —x is the deviation between x; and the mean. We prove first that the sum of these deviations is 0, using the foregoing definition of x:

n n n n y^ (xi — x) = ^^ xi ~ ^^ * = ^^ X{ — nx — nx — nx = 0

Furthermore, n n n n n y^(xj - x)2 = ^(x2 - 2xxi -+- x2) = ^^xf ~ 2Jc ^xi + y^j2

Afoie: We have considered some useful algebraic properties of sums. A frequent error is a failure to observe that, in general,

It is important to note that the sum of the squares is not generally equal to the square of the sum. For example, Yll=\ x} = xf + x2 whereas x,)2 = (x4- x2)2 = x2 4- 2.xix2 + x2, so the two are equal iff x\x2 = 0—that is, x\ or x2 (or both) must be zero. More generally, so the sum of the cross products is not equal to the products of the individual sums.

### Useful Formulas

If you asked a group of 10-12-year-old students to sum all the numbers from 1 to 100, would you expect to have a correct answer within 1 hour? According to reliable sources, Carl F. Gauss solved a similar problem in his tenth year. His teacher asked his students to sum 81,297 + 81,495 + 81,693 + - • - + 100,899. There are 100 terms and the difference between successive terms is constant and equal to 198. Obviously, the teacher chose this sum knowing that a trick could yield the answer quickly. Thus, the laboriously derived answers of the students could easily be checked. But Gauss, who later became one of the world's leading mathematicians, gave the right answer, which is 9,109,800, in only a few minutes.

Applied to the easier problem of finding the sum 1 + 2 + • • • + n, Gauss' argument was probably as follows: First, write the sum x in two ways

2x = (1 + n) + [2 + (n - 1)] + • • • + [(« - 1) + 2] + (n + 1)

Thus, we have the result:

x = 1+2 +----h(n- l) + /i x = n + (n — \) ----+ 2+1

Summing vertically gives n i=i

The following two summation formulas are sometimes useful:

^ r = l2 + 22 + 32 + ■ • ■ + n2 = i«(n + l)(2n + 1) [B.6]

Check to see if these formulas are true for n = 1, 2, and 3. One way of proving that they are valid generally is to use mathematical induction, as discussed in Section B.5.

Newton's Binomial Formula

We all know that (a + b)1 = a + b and (a + b)2 = a2 + + b2. Using the latter equality and writing (a + b)3 = (a + ¿>)2(a + b) and (a + b)4 = (a + b)2(a + b)2, we find that

What is the corresponding formula for (a + b)m, where m is an arbitrary positive integer? The answer is given by the Newton binomial formula:

(a + b)m =am+ (™)am~lb + ("V-2*2 + " " '

Here the binomial coefficient

U; = ki as explained in Section 7.4. Formula [B.8] is proved in Section 7.4. In general, (7) = m and (J) = 1. For m = 5, we have

(a + b)5 = a5 + 5 a4b + 10 a2b2 + \0a2b3 + 5 ab4 + b5

If we study the coefficients in the expansions for the successive powers of (a + b), we have the following pattern, called Pascal's triangle (though it was actually known in China by around 1100, long before Blaise Pascal was bom):

1 1 1 2 1 13 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1

This table can be continued indefinitely. The numbers in this triangle are the binomial coefficients. For instance, the numbers in row 6 (when the first row is numbered 0) are

0 0