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In the sequence 1, 2, 4, 8, 16, 32, …,

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In the sequence 1, 2, 4, 8, 16, 32, …,

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In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term. What is the sum of the 16th, 17th, and 18th terms in the sequence?
$$A.\ \ \ 2^{18}$$ $$B.\ \ \ 3\left(2\right)^{^{17}}$$ $$\left(C.\ \ \ 7(2\right)^{16}$$ $$D.3\left(2\right)^{16}$$ $$E.7\left(2\right)^{15}$$

The OA is E.

What is the formula that I should use here? Experts, can you show me how to solve this PS question? Thanks in advanced.

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Hello VJesus12.

Let's take a look at your question.

We can rewrite the sequence as follows: $$2^0\ \ ,\ \ \ 2^1\ ,\ \ 2^2\ ,\ 2^3\ ,\ \ 2^4\ ,\ \ \ 2^5\ ,\ \ \ 2^6\ ,\ \dots$$ This is equivalent to say that the nth term is $$a_n=2^{n-1}.$$ So, we have that $$a_{16}=2^{15},\ \ a_{17}=2^{16}\ \ \ and\ a_{18}=2^{17}.$$ Therefore $$a_{16}+a_{17}+a_{18}=2^{15}+2^{16}+2^{17}=2^{15}\left(1+2+2^2\right)=7\left(2\right)^{15}.$$ So, the correct answer is E.

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.

Regards.

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VJesus12 wrote:
In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term. What is the sum of the 16th, 17th, and 18th terms in the sequence?
$$A.\ \ \ 2^{18}$$ $$B.\ \ \ 3\left(2\right)^{^{17}}$$ $$\left(C.\ \ \ 7(2\right)^{16}$$ $$D.3\left(2\right)^{16}$$ $$E.7\left(2\right)^{15}$$

The OA is E.

What is the formula that I should use here? Experts, can you show me how to solve this PS question? Thanks in advanced.
the series is 1,2,4,8,16,32,............

in terms of power of 2 we can rewrite it in the form

2^0,2^1,2^2,2^3,2^4,2^5,........

=> nth term = 2^(n-1)

therefore 16th, 17th ,and 18th term would be 2^15,2^16,and 2^17

and their sum = 2^15+2^16+2^17= 2^15 * (1+2+4) = 7(2^15)

hence option E is correct

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VJesus12 wrote:
In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term. What is the sum of the 16th, 17th, and 18th terms in the sequence?
$$A.\ \ \ 2^{18}$$ $$B.\ \ \ 3\left(2\right)^{^{17}}$$ $$\left(C.\ \ \ 7(2\right)^{16}$$ $$D.3\left(2\right)^{16}$$ $$E.7\left(2\right)^{15}$$
First notice the PATTERN:
term_1 = 1 (aka 2^0)
term_2 = 2 (aka 2^1)
term_3 = 4 (aka 2^2)
term_4 = 8 (aka 2^3)
term_5 = 16 (aka 2^4)
.
.
.
Notice that the exponent is 1 LESS THAN the term number.

So, term_16 = 2^15
term_17 = 2^16
term_18 = 2^17

We want to find the sum 2^15 + 2^16 + 2^17
We can do some factoring: 2^15 + 2^16 + 2^17 = 2^15(1 + 2^1 + 2^2)
= 2^15(1 + 2 + 4)
= 2^15(7)
= E

Cheers,
Brent

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