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If $a_{1}+a_{2}+a_{3}+\dots+a_{n}=3(2^{n+1}-2)$, for every $n\geq 1$, then $a_{11}$ equals ______

Given that, $a_{1} + a_{2} + a_{3} + \dots + a_{n} = 3(2^{n+1} – 2)$

Now, put $n=11,$ we get

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10}+a_{11} = 3(2^{12}-2)$

$= 3 (4096-2)$

$= 3 \times 4094$

$= 12282 \quad \longrightarrow (1)$

And put $n=10,$ we get

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10} = 3(2^{11}-2)$

$= 3 (2048 – 2)$

$= 3 \times 2046$

$= 6138 \quad \longrightarrow (2)$

From  the equation $(1),$ subtract the equation $(2),$ we get

$a_{11} = 12282 – 6138$

$\boxed{a_{11} = 6144}$

Correct Answer $: 6144$
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