Question

Evaluate the summation of 2 times negative 2 to the n minus 1 power, from n equals 1 to 7..

Answer 1

\[\sum_{n=1}^{7} 2(-2)^n\]

Answer 2

the last n should be ^n-1 not just n

Answer 3

−256

−84

84

86

Answer 4

\[\large \sum_{n=1}^{7} 2(-2)^{n-1}\]
?

Answer 5

yes

Answer 6

Okay for n=1 the exponent on -2 is 0. So that factor is?

Answer 7

how do I find the factor

Answer 8

\[ (-2)^0 = 1\]

any number to the zero power is 1 (but not defined for 0 to the 0).

So the first term is 2(1).

Answer 9

So far, the series is 1 + ...

Now for the n=2, term.

Answer 10

ok1 +2 ?

Answer 11

When n=2, what is:
\[\large 2(-2)^{n-1}\]

Answer 12

=4 -4 ^n-1

Answer 13

For n=2.

\[\large 2(-2)^{2-1} = 2(-2)^1 = 2(-2) = -4\]

So the sum so far is 1 + (-4) + ...

Now what if n=3?