He barely tinished stating the problem when young Carl came forward and

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1 .. Figurate numbers (a) Figure.. " He barely tinished stating the problem when young Carl came forward and placed his slate on the teacher's desk, void of calculation, with the correct answer: 5050, When asked to explain, Gauss admitted he recognized the pattern 00 = 0, 99 = 0, 98 = 0, and so on to 50 5 = 0. Since there are tifty such pairs, the sum must be 50 0 = The pattern for the sum (adding the largest number to the smallest, the second largest to the second smallest, and so on) is i llustrated in Figure., where the rows of balls represent positive integers. The number tn =... n for a positive integer n is called the nth triangular number, from the pattern of the dots on the left in Figure.. Young Carl correctly computed tloo = However, this solution works only for n even, so we tirst prove Theorem LI. For al! n ~ l, tn = n(n )/. Prao! We find a pattern that works for any n by arranging two copies of tn to form a rectangular array of balls in n rows and n columns. Then we have tn = n(n ), or tn = n(n )/. See Figure.. Proo! We give two connhi'lallll. in two ways, fi rst as a square in each L-shaped region of one-to one correspondence triangular array of balls in array of balls. The same idea can be lowing sequence of identities: Each row begins VI can be proved by j In Figure.4, VI number of sm ail cubes in the pi =. 9 Fig ure.. The counting procedure in the preceding combinatorial proof is double counting or the Fubini principle, as mentioned in the Introduction. We employ the same procedure to prove that su ms of odd numbers are squares. Theorem.. For al! n ~, 5... (n - ) = n.

2 .. Figurate numbers (a) Figure.. Proof We give two combinatorial proofs. In Figure.a, we count the balls in two ways, first as a square array of balls, and then by the number of balls in each L-, haped region of similarly colored balls. In Figure.b, we see a one-to one correspondence (illustrated by the color of the balls) between a triangular array of balls in rows with l,, S,..., n - balls, and a square array of balls. The same idea can be employed in three dimensions to establish the fo l lowing sequence of identities: =, = 7 8, 9 0 Il = 4 5, etc. Each row begins with a square. The general pattern n (n )... (n n) = ( n ) '" (n n) can be proved by induction, but the following visual proof is nicer. In Figure la, we see the n = 4 version of the identity where counting the number of sm ail cubes in the pile in two different ways yields = 4. Figure.4.

3 6,.. Sums of sq uares, triangular nun Hence (... n ) = (: the result follows. In the second proof, we write each 5 those numbers in a triangular array, cre triangulai' array by 0 and 40, am triangular array. See Figure.0 [Kung. Figure.8. The arrows denote the correspondence between an element of the set with ln elements and a pair of elements from a set of n elements.. Sums of squares, triangular numbers, and cubes Having examined triangular numbers and squares as sums of integers and sums of odd inlegers. we now consider sums of triangular numbers and sums of squares. n(n )(n ) Tbcorem.7. For al! n ::::,... n = 6. Proo! We gi e two proofs. In the orst we exhibit a one-to-one corresponden ce between three copies of... n and a rectangle whose dimensions are n and... n = n(n ) / [Gardner. ]97J. See Figure.9. m Em EB I-t--t--i 0 gmemebd [ n // TI // n-i n - [ - n \ n n Il Il - m /tltl //ttl // Theorem.8. For all n ::: l, Il T Proo! n Figure., we stack layers gular numbers. The sum of the which is the same as the total volume we "slice off" small pyramids (shaderl on the top of the cube from which il gular pyramid minus sorne smaller. of the base. T hus II... ln = -(n 6 IIlEffil EBD Fig ure.9.

4 .. Su ms of squares, triangular numbe rs, and cu bes 7 Hcnce (... n ) = (n )(... n) from which the rcsult fo llows. In the second proof, we write each. guare k as a sum of k ks, then place those numb rs in a tri angular array, cr ale two more arrays by rotating the triangulai' array by 0 0 and 40 0, and add corresponding entri s in each triangul ar array. Sc Figure. 0 [Kung, 989]. n n n n- ] n-l n n-] n- n- n-l n n-i Il- j.. n-l n '''n- n n Il n ". n n Il -]... - n n n l nn ] n nl ". Figure.0. n( l)(n ) Theorem.8. For au n :::, II... ln = 6. Prao! ln Figure., we stack layers of unit cubes to represent th triangular numbers. The sum of the triangulai' numb 'S is total numb r of cubes, which is th same as the total volume of the cubes. To compute the volume, we "slice If" small pyramids (shaded gray) and place each small pyramid on the top o r the cube from which il came. The result is a large right triangulai' pyramid minus some smaller right triangular pyramids along one edge of the base. n(n )( ) ThuSII t " ' ln =-(n l ) - ( ) () \ Figure..

5 8.. There are infinitely many primes In the proof we evaluated the sum of the first n triangular numbers by computing volumes of pyramids. This is actually an extension of the Fubini principle from simple enumeration of objects to additive measures such as length, area and volume. The volume version of the Fubini principle is: computing the volume ofan object in two different ways yields the same number; and similarly for length and area. We cannot, however, ex tend the Cantor principle to additive measures-for example, one can construct a one-toone correspondence between the points on two line segments with different lengths. Tbeorem.9. Foralln::::., n = (/ IY = t;. Proo! Again, we give two proofs. In the first, we represent k as k copies of a square with area k to establish the identity rcupillari, 989; Lushbaugh, 965J. Il 4 6 n 6 9 / n n n n Figure. We conclude this section with a theoret sum of integers. Tbeorem.] O. For all n :::, [:'= [ j. Prao! We represent the double sum as a pute the volume of a rectangular box corn tion. See Figure.4. r ~ ; f- - ~- It!- -mi-.."""'-- _. Fi gure.. r- --- In Figure l., we have 4(.... ) = [n( )j (for n = 4). For the second proof, we use the fact that... (n - ) n (n - )... = n (see Challenge. a) and consider a square array of numbers in which the element in row i and column j is ij, and sum the numbers in two different ways [Pouryoussefi, 989]. Summing by columns yields L7= i (L7= i)... n (L7 = i) = (I:7=i), while summing by the L-shaped shaded regions yields (using the result of Cha llenge!.la)..... n n = L7= i. Figure,: Two copies of the sum 5 = I:7= I:j= box with base n and height n, hence CI two ways yields 5 = n, or S = /.. There are infinitely m Reductio ad absurdu a mathematician's thcm any chess gamb of a pawn or evell Il game. The earliest proof that there are infinitel) in the Elements (Book IX, Proposition