Page 101 - Numbertheory
P. 101

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2 + 4 + 6 + ... + 2n

2 + 4 + 6 + ... + 2n = 2(1+ 2 + 3 + ... + n )

=      n  (n +  1)
   2       2     

= n(n +1)

. n ≥ 1 1+ 3 + 5 + ... + (2n −1) = n 2         ()

1+ 2 + 3 + 4 + ... + (2n −1) + 2n
= [1+ 3 + ... + (2n −1)] + [2 + 4 + 6 + ... + 2n]
= [1+ 3 + ... + (2n −1)] + 2[1+ 2 + 3 + ... + n]

             2n (2n +1) = [1+ 3 + ... + (2n −1)] + 2 n (n +1)
                   22

           .1+ 3 + ... + (2n −1) = 2n 2 + n − n 2 − n = n 2

        105 ( )

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