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The number of distinct terms in the expansion of (1 + x + x³)¹⁰ is:

1. 11
2. 29
3. 30
4. 31

Correct answer: 30

Solution

By multinomial theorem, possible exponents of x are of the form q+3r with p+q+r=10. This gives n = 10-p+2r. Every integer from 0 to 28 and also 30 is achievable, but 29 is not.

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