Question

Use sigma notation to represent the following series for the first 10 terms. 3 + (-6) + 12 + (-24) + ⋯

Answer 1

`$$\sum_(n=1)^(10) -3(-1)^(n) \hspace{3}2^(n-1) $$`

`n\in N`

Explanation:

First, as you can see, the sequence is alternating its sign everytime, so you can deduce this term:

`(-1)^(n)`

The sequence start from 3, and the following terms are the product between 3 and
`2^(n-1)`. Take a look:

For n=1

`3*2^(1-1) =3*2^(0) =3*1=3`

For n=2

`3*2^(2-1) =3*2^(1) =3*2=6`

And so on...

Let's verify the formula including all terms:

Since the sequence start from 3, we must change the 3 for -3. Because
`(-1)^(n)` is always negative for the first term:

For n=1

`(-1)^(1) \hspace{3}-3*2^(1-1) =(-1)-3*2^(0) =(-1)-3*1=3`

For n=2

`(-1)^(2) \hspace{3}-3*2^(2-1) =(1)-3*2^(1) =(1)-3*2=-6`

For n=3

`(-1)^(3) \hspace{3}-3*2^(3-1) =(-1)-3*2^(2) =(-1)-3*4=12`

For n=4

`(-1)^(4) \hspace{3}-3*2^(4-1) =(1)-3*2^(3) =(1)-3*8=-24`

So, one possible sequence is:

`a_n=-3(-1)^(n) \hspace{3}2^(n-1)`

And the serie would be given by:

`$$\sum_(n=1)^(10) a_n $$ = $$\sum_(n=1)^(10) -3(-1)^(n) \hspace{3}2^(n-1) $$`