Sum
Without expanding, find the value of (x + 1)4 − 4(x + 1)3 (x − 1) + 6 (x + 1)2 (x − 1)2 − 4(x + 1) (x − 1)3 + (x − 1)4
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Solution
Let x + 1 = a and x − 1 = b
We notice that 1, 4, 6, 4, 1 are the values of 4C0, 4C1, 4C2, 4C3, 4C4 respectively and
∴ the given expression becomes
4C0a4b0 − 4C1a3b + 4C2a2b2 − 4C3ab3 + 4C4a0b4
= (a − b)4
= [x + 1 − (x − 1)]4
= (x + 1 − x + 1)4
= 24
= 16
Concept: Binomial Theorem for Positive Integral Index
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