12 SPECIAL TOPICS IN THEORETICAL ARITHMETIC

Th. Thus the sum of the first n odd numbers starting with
Tas

33. Consider the series of squares of the natural numbers

12’ 22, 32’ s n2
Th. Herewewill prove that.S, = n(n + 1)(2n + 1)/6, that s,
that S1=1X2X3/6
Se=2X3X5/6
=3X4X17/6

S3

Se = k(k + )2k + 1)/6
Skt1 = ( + 1)(k + 2)(2k + 3)/6

Of these statements the first is evidently tru.
And, if the kth is tru, that is, if

Se = k(k + 1)(2k + 1)/6,
then Sk+1 Sk + Ar41 = k(k -|— 1)(2]3 + 1)/6 + (k + 1)2
(k + 1)[k(2k + 1) + 6(k + 1)]/6
(k + D2k + 7k + 6]/6
(B + 1)(k + 2)(2k + 3)/6
( + 1)(k + 2)(2k + 3)/6

Hence the kth statement in the series implies the next.
Therfor the statements are all tru and

Therfor Siy1

Sn = n(n + 1)(2n + 1)/6 for all values of n.
Thus 124224+ 32 4 -+ + 22 = n(n + 1)2n + 1)/6
34. Th. We can similarly prove that
P+2R 3+ et = (n(n + 1)/2) -

Hence the sum of the cubes of the first n natural numbers is
equal to the square of the sum of these natural numbers.