Question

the summation of 3 times negative 2 to the n minus 1 power, from n equals 1 to 5..

Answer 1

well its

\[\sum_{n = 1}^{5} 3 \times (-2)^{n - 1} = 3 \times (-2)^{1 - 1} + 3\times (-2)^{2 -1} + 3 \times (-2)^{3 -1} + 3 \times (-2)^{4 -1} + 3 \times (-2)^{5 - 1}\]

now just calculate it out

Answer 2

oops last term is missing its

\[+ 3 \times (-2)^{5 - 1}\]

Answer 3

93?