Each term in the sequence below is three times the previous term. What is the product of the 100th and 101st terms of the sequence?

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Every term in the sequence is 3 raised to a power. The first term, 1, is 3 0 . The second term, 3, is 3 1 . The value of the exponent is one less than the position of the term in the sequence. The 100th term of the sequence is equal to 3 (100 – 1) = 3 99 and the 101st term in the sequence is equal to 3 (99 + 1) = 3 100 . To multiply two terms with common bases, add the exponents of the terms: (3 99 )(3 100 ) = 3 199 .

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