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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