Page 260 - aljaber

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2k + 1
( )S2k+1 2  + (2k)× 1
= 2 k2 +1

= 2k + 1 ( 2k 2 + 2k + 2)
2
= (2k + 1)(k2 + k + 1)

= 2k 3 + 3k2 + 3k + 1

= k 3 + 3k 2 + 3k + 1 + k 3 = (k + 1)3 + k 3

:( ) ()

S3n − S2n − Sn

= 3n (2a + (3n − 1)d ) − 2n (2a + (2n − 1)d )

22

−n (2a + (n − 1)d )

2

= n  6a + 9nd − 3d − 4a − 4nd + 2d − 2a − nd + d 
2

= n  4nd  = 2n2d
2