In the sequence S, each term after the first is twice the previous term. If the first term of sequence S is 3, what is the sum of the 14th, 15th, and 16th terms in sequence S?
(A) 3(2^16)
(B) 9(2^15)
(C) 21(2^14)
(D) 9(2^14)
(E) 21(2^13)
OA E
Source: Manhattan Prep
In the sequence S, each term after the first is twice the
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\[S \to \,\,\,{a_{\,1}}\,,\,\,{a_{\,2}}\,,\,\,{a_{\,3}}\,,\,\,\, \ldots \,\,\,\,\left\{ \begin{gathered}BTGmoderatorDC wrote:In the sequence S, each term after the first is twice the previous term. If the first term of sequence S is 3, what is the sum of the 14th, 15th, and 16th terms in sequence S?
(A) 3(2^16)
(B) 9(2^15)
(C) 21(2^14)
(D) 9(2^14)
(E) 21(2^13)
Source: Manhattan Prep
\,{a_{\,1}} = 3 \hfill \\
{a_{\,n}} = 2 \cdot {a_{\,n - 1}}\,\,\,\,\left( {n \geqslant 2} \right) \hfill \\
\end{gathered} \right.\]
\[? = {a_{\,14}} + {a_{\,15}} + {a_{\,16}}\]
The PATTERN is clear:
\[{a_{\,\boxed2}} = {2^{\,\boxed1}} \cdot {a_{\,1}}\,\,\,\,\,\,;\,\,\,\,\,\,{a_{\,\boxed3}} = 2 \cdot {a_{\,2}} = {2^{\,\boxed2}} \cdot {a_{\,1}}\,\,\,\,\,\,;\,\,\,\,\,\,{a_{\,\boxed4}} = 2 \cdot {a_{\,3}} = {2^{\,\boxed3}} \cdot {a_{\,1}}\,\,\,\,\,\,;\,\,\,\,\,\, \ldots \]
\[\left. \begin{gathered}
{a_{\,14}} = {2^{\,13}} \cdot {a_{\,1}} \hfill \\
{a_{\,15}} = 2 \cdot {2^{\,13}} \cdot {a_{\,1}}\,\, \hfill \\
{a_{\,16}} = {2^2} \cdot {2^{\,13}} \cdot {a_{\,1}} \hfill \\
\end{gathered} \right\}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,?\,\,\, = \,\,\,{2^{13}} \cdot {a_1} \cdot \left( {1 + 2 + {2^{\,2}}} \right)\,\,\,\mathop = \limits^{{a_{\,1}}\, = \,\,3} \,\,\,\,3 \cdot 7 \cdot {2^{\,13}}\]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
fskilnik.
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The first term is 3, the second is 2 x 3, the third is 2^2 x 3, and so on. We see that the general formula for the nth term is: a_n = 2^(n - 1) x 3. Thus, the 14th term is 2^13 x 3, the 15th term is 2^14 x 3, and the 16th term is 2^15 x 3, so the sum is:BTGmoderatorDC wrote:In the sequence S, each term after the first is twice the previous term. If the first term of sequence S is 3, what is the sum of the 14th, 15th, and 16th terms in sequence S?
(A) 3(2^16)
(B) 9(2^15)
(C) 21(2^14)
(D) 9(2^14)
(E) 21(2^13)
OA E
Source: Manhattan Prep
2^13 x 3 + 2^14 x 3 + 2^15 x 3
We see that the common factor for all three terms is 2^13 x 3, which we factor from each term:
2^13 x 3(1 + 2 + 2^2)
2^13 x 3(7) = 2^13 x 21
Answer: E
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term1 = 3BTGmoderatorDC wrote:In the sequence S, each term after the first is twice the previous term. If the first term of sequence S is 3, what is the sum of the 14th, 15th, and 16th terms in sequence S?
(A) 3(2^16)
(B) 9(2^15)
(C) 21(2^14)
(D) 9(2^14)
(E) 21(2^13)
OA E
Source: Manhattan Prep
term2 = (3)(2) = (3)(2¹)
term3 = (3)(2)(2) = (3)(2²)
term4 = (3)(2)(2)(2) = (3)(2³)
term5 = (3)(2)(2)(2)(2) = (3)(2�)
.
.
term14 = (3)(2^13)
term15 = (3)(2^14)
term16 = (3)(2^15)
What is the sum of the 14th, 15th, and 16th terms in sequence S?
Sum = (3)(2^13) + (3)(2^14) + (3)(2^15)
Factor out 3(2^13) to get: (3)(2^13)[1 + 2 + 2^2]
Simplify to get: (3)(2^13)[7]
Rewrite as: (21)(2^13)
Answer: E
Cheers,
Brent