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Question 18.

If a1+a2+a3+....+an=3(2n+1−2)a_1 + a_2 + a_3 + .... + a_n = 3(2^{n + 1} - 2), for every n≥1n \geq 1, then a11a_{11} equals

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11th term of series = a11a_{11}​ = Sum of 11 terms - Sum of 10 terms = 3(211+1−2)−3(210+1−2)3(2^{11+1}−2)-3(2^{10+1}−2)

= 3(212−2−211+2)=3(211)(2−1)=3×2113(2^{12}−2−2^{11}+2) = 3(2^{11})(2-1) = 3 \times 2^{11} = 6144

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