Q12 of 25 Page 8

Find (x + 1)6 + (x - 1)6. Hence or otherwise evaluate


(x + 1)6 + (x - 1)6
= [6C0 x6 + 6C1 x5 + 6C2 x4] + [6C3 x3 + 6C4 x2 + 6C5 x1 - 6C6] +
[6C0 x6 - 6C1 x5 + 6C2 x4 - 6C3 x3 + 6C4 x2 + 6C5x1 – 6C6]
= 2[6C0x6 + 6C2 x4 + 6C4 x2 + 6C6]
= 2[x6+ 15x4+ 15x2+ 1]
Thus, (+ 1)6 + (- 1)6
= 2 [()6 + 15()4 + 15 () + 1]
= 2 [8+ 15 (4)+ 15 (2) + 1]
= 2 [8 + 60 + 30+ 1]= 198.

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