If a_1+a_2+a_3+...+a_n=3(2^{n+1}-2), for every n ≥ 1, then a_11 equals

CAT 2019 Shift 1 Paper

Question : 69 of 100

Marks: +1, -0

a_1+a_2+a_3+...+a_n=3(2^{n+1}-2),

for every n ≥ 1, then

a_11

equals

Solution:

n=1, a_{1}=3(2^{1+1}-2)=6=3 × 2^{1}

{n}=2, a_{1}+a_{2}=3(2^{2+1}-2)=18 ⇒ a_{2}=18-a_{1}=12=3 × 2^{2}

{n}=3,a_{1}+a_{2}+a_{3}=3(2^{3+1}-2)=42 ⇒ a_{3}=42-(a_{1}+a_{2})=24=3 × 2^{3}

Following the pattern,

a_{n}=3 × 2^{n}

Therefore,

a_{11}=3 × 2^{11}=6144

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