Question

Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15.

n an
1 4
2 −12
3 36

Answer 1

1. We notice that the signs of the terms in the sequence alternative, first positive then negative, and so on.
2. We notice that all of the terms are divisible by 4.
3. When we divide the terms by 4, and ignore the signs for the moment, we get 1, 3 and 9. We notice that these are powers of 3 (1 being 3 to the power zero).

\[4 = (-1)^{1+1}4(3^{1-1})\]
\[-12 = (-1)^{2+1}4(3^{2-1})\]
\[36 = (-1)^{3+1}4(3^{3-1})\]

The pieces of the expressions involving the -1s are just there to get the alternating signs. By constructing these systematically we can see that a pattern has emerged. We need only copy it into a summation expression.
\[\sum_{k=11}^{16}(-1)^{k+1}4(3^{k-1})\]

Answer 2

Since the constant 4 appears in every term we could put it outside the summation. Otherwise we've finished.