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If a1 + a2 + a3 + … + an = 3(2^n + 1 − 2 ), for every n ≥ 1, then a11

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If a1 + a2 + a3 + … + an = 3(2^n + 1 − 2 ), for every n ≥ 1, then a11  [#permalink]

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New post 02 Apr 2020, 02:04
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  55% (hard)

Question Stats:

69% (02:20) correct 31% (03:02) wrong based on 42 sessions

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If a1 + a2 + a3 + … + an = 3(2^n + 1 − 2 ), for every n ≥ 1, then a11  [#permalink]

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New post 02 Apr 2020, 02:18
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If \(a_1 + a_2 + a_3 + … + a_n = 3(2^{n + 1} − 2 )\), for every \(n ≥ 1\), then \(a_{11}\) equals

If \(n = 10\), then
—>\( a_1+ ...+a_{10} = 3( 2^{11} —2)\).

If \(n= 11\), then
—>\( a_1+...a_{10} + a_{11} = 3(2^{12} —2)\)

Well, subtracting 2nd(n=11) statement from 1st (n=10):
—> \(a_{11} = 3(2^{12} —2) —3(2^{11}—2)\) =
\( 3( 2^{12} —2^{11} ) = 3*2^{11}( 2–1) = 3* 2048 = 6144\)

Answer (E)

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Re: If a1 + a2 + a3 + … + an = 3(2^n + 1 − 2 ), for every n ≥ 1, then a11  [#permalink]

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New post 02 Apr 2020, 03:16
1
Bunuel wrote:
If \(a_1 + a_2 + a_3 + … + a_n = 3(2^{n + 1} − 2 )\), for every \(n ≥ 1\), then \(a_{11}\) equals

A. 1536
B. 2012
C. 2048
D. 3072
E. 6144


\(a_1 + a_2 + a_3 + … + a_n = 3(2^{n + 1} − 2 )\)

Substitute, n = 10
--> \(a_1 + a_2 + a_3 + … + a_{10} = 3(2^{10 + 1} − 2 )\)
--> \(a_1 + a_2 + a_3 + … + a_{10} = 3(2^{11} − 2)\) ........ (1)

Substitute, n = 11
--> \(a_1 + a_2 + a_3 + … + a_{10}+ a_{11} = 3(2^{11 + 1} − 2 )\)
--> \(a_1 + a_2 + a_3 + … + a_{10}+ a_{11} = 3(2^{12} − 2 )\) ........ (2)

(1) - (2) gives,
--> \((a_1 + a_2 + a_3 + … + a_{10} + a_{11}) - (a_1 + a_2 + a_3 + … + a_{10}) = 3(2^{12} − 2) - 3(2^{11} − 2)\) ........ (2)
--> \(a_{11} = 3(2^{12} − 2^{11}) = 3*2048= 6144\)

Option E
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Re: If a1 + a2 + a3 + … + an = 3(2^n + 1 − 2 ), for every n ≥ 1, then a11  [#permalink]

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New post 02 Apr 2020, 07:24
1
a1+a2+a3+…+an=3(2^n+1−2), for every n≥1,
or, a1+a2+a3+…+an=3*2(2^n-1)
a1 = 3*2 (2^1-1)
a1+a2 = 3*2(2^2-1)
=> a2 = 3*2(2^2-2^1)

therefore, a11 = 3*2(2^11-2^10)
or, a11 = 6 *1024 = 6144

correct answer is E
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If a1 + a2 + a3 + … + an = 3(2^n + 1 − 2 ), for every n ≥ 1, then a11  [#permalink]

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New post 03 Apr 2020, 08:03
Bunuel wrote:
If \(a_1 + a_2 + a_3 + … + a_n = 3(2^{n + 1} − 2 )\), for every \(n ≥ 1\), then \(a_{11}\) equals

A. 1536
B. 2012
C. 2048
D. 3072
E. 6144

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Asked: If \(a_1 + a_2 + a_3 + … + a_n = 3(2^{n + 1} − 2 )\), for every \(n ≥ 1\), then \(a_{11}\) equals


\(a_1 + a_2 + a_3 + … + a_{10} = 3(2^{10 + 1} − 2 )=3(2^{11}-2) \)
\(a_1 + a_2 + a_3 + … + a_{11} = 3(2^{11 + 1} − 2 )=3(2^{12}-2) \)
\(a_{11} = 3(2^{12}-2) - 3(2^{11}-2) = 3(2^{12} - 2^{11}) = 3*2^{11} = 3*2048 = 6144\)

IMO E
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If a1 + a2 + a3 + … + an = 3(2^n + 1 − 2 ), for every n ≥ 1, then a11   [#permalink] 03 Apr 2020, 08:03

If a1 + a2 + a3 + … + an = 3(2^n + 1 − 2 ), for every n ≥ 1, then a11

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