Given:
\(a_1=2^0=1\);
\(a_2=2^1=2\);
\(a_3=2^2=4\);
...
\(a_n=2^{n-1}\);

Thus \(a_{16}+a_{17}+a_{18}=2^{15}+2^{16}+2^{17}=2^{15}(1+2+4)=7*2^{15}\).

So you don't actually need geometric series formula.

Answer: E.

But still if you are interested:

Sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}-1)}{r-1}\), where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\).

Sum of infinite geometric progression with common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term.

Hope it helps.
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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
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Given:
\(a_1=2^0=1\);
\(a_2=2^1=2\);
\(a_3=2^2=4\);
...
\(a_n=2^{n-1}\);

Thus \(a_{16}+a_{17}+a_{18}=2^{15}+2^{16}+2^{17}=2^{15}(1+2+4)=7*2^{15}\).

So you don't actually need geometric series formula.

Thanks very Much! This is an excellent approach.

16th term = 2^15 (since 2^0 = 1). Hence we need 2^15+2^16+2^17.

Now take smaller numbers: 2²+2³+2^4 = 28 = 7*(2²) (which is the first term), hence 7*(2^15) will be right. E.

I don't understand where the 2^16 and 2^17 go. and why is a16 + a17 + a18 = 2^15 + 2^16 + 2^17

Note : Sorry I can't do the subscripts for the a's

1st term: \(1 = 2^0\)
2nd term: \(2^1\)
3rd term: \(2^2\)
4th term: \(2^3\)
5th term: \(2^4\)

So looking at the pattern, what will be the 16th term? It will be \(2^{15}\)
What about the 17th term? \(2^{16}\)
What about the 18th term? \(2^{17}\)

When you add them, you get \(2^{15} + 2^{16} + 2^{17}\)
Now you take \(2^{15}\) common from the 3 terms. You are left with

\(2^{15}* (1 + 2 + 2^2) = 2^{15}*7\)

Note that \(2^{16}\) has 16 2s. When you take out 15 2s, you are left with a single 2. Similarly, \(2^{17}\) has 17 2s. When you take out 15 2s, you are left with two 2s i.e. \(2^2\)
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Pattern :
1st term, --------------------------------------, 6th term ,...
can be written as :
1 ,(2) ,(2*2),(2*2*2),(2*2*2*2) , (2*2*2*2*2),...

which again can be written as :
1 , 2^1, 2^2 , 2^3 , 2^4 , 2^5 ,...

Therefore ,
16th term : 2^15 ---(1)
17th term : 2^16 ---(2)
18th term : 2^17 ---(3)

Adding (1),(2) & (3)
2^15 + 2^16 + 2^17 = 2^15(1+ 2^1 + 2^2) = 2^15 ( 1+2+4) = 2^15 (7)

Ans : E

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