Write the sum using sigma notation.

-3 + 0.3 – 0.03 + 0.003 – 0.0003

Answer

Let's deconstruct this series. First, we can pull out a factor of 3 from each term to get

-3 + 3(0.1) – 3(0.01) + 3(0.001) – 3(.0001).

The things in the parentheses are are powers of .1, so write them that way:

-3(0.1)^{0} + 3(0.1)^{1} – 3(0.1)^{2} + 3(0.1)^{3} – 3(0.1)^{4}

It makes sense to use *n* = 0 as the starting index, so to get the signs correct we need to have a factor of (-1)^{n + 1}:

(-1)^{1} × 3(.1)^{0} + (-1)^{2} × 3(.1)^{1} + (-1)^{3} × 3(.1)^{2} + (-1)^{4} × 3(.1)^{3} + (-1)^{5} × 3(.1)^{4}

Finally, we put things into sigma notation:

It would also make sense to start at *n* = 1, which is the exponent on the first factor of -1. Then the series looks like

(-1)^{1} × 3(0.1)^{0} + (-1)^{2} × 3(0.1)^{1} + (-1)^{3} × 3(0.1)^{2} + (-1)^{4} × 3(0.1)^{3} + (-1)^{5} × 3(0.1)^{4}

and we would write it in sigma notation as