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Exercise 8.1 · Q9

Q.Find the indicated term in the sequence whose nnth term is an=(−1)n−1 n3a_n = (-1)^{n-1}\,n^3; a9a_9.

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Substitute n=9n = 9 into the formula an=(−1)n−1 n3a_n = (-1)^{n-1}\,n^3 and evaluate the sign and cube separately. The ninth term is 729\boxed{729}.

Understanding the Formula

A sequence is defined by its general term ana_n, which tells you how to compute any term from its position number nn. Here we have an=(−1)n−1 n3a_n = (-1)^{n-1}\,n^3, which is the product of two parts:

  • The factor (−1)n−1(-1)^{n-1} controls the sign and creates an alternating pattern
  • The factor n3n^3 gives the magnitude as the cube of the position

The exponent n−1n-1 in the sign factor is key. When nn is odd, n−1n-1 is even, so (−1)n−1=+1(-1)^{n-1} = +1. When nn is even, n−1n-1 is odd, so (−1)n−1=−1(-1)^{n-1} = -1. This means odd-positioned terms are positive and even-positioned terms are negative.

Tip

To remember the alternating pattern: (−1)n−1(-1)^{n-1} starts positive (at n=1n=1), while (−1)n(-1)^n starts negative. The choice depends on whether the sequence should begin with a positive or negative term.

Finding a9a_9

We need the ninth term, so we substitute n=9n = 9 into the formula.

  1. Evaluate the sign factor: (−1)9−1=(−1)8=+1(-1)^{9-1} = (-1)^8 = +1 …

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