Question

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

Answer 1

Answer:

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

$n\in N$

Step-by-step 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:

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