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

Entered answer:

Solution

✅ Correct Answer: 6144

If we know the sum of first nn terms and the sum of first (n−1)(n-1) terms, then:

an=Sn−Sn−1a_n = S_n - S_{n-1}

This is because when we subtract the sum of first 10 terms from the sum of first 11 terms, we get just the 11th term!


Using our formula with n=11n = 11:

S11=3×(211+1−2)S_{11} = 3 \times (2^{11+1} - 2)

=3×(212−2) = 3 \times (2^{12} - 2)

=3×(4096−2)= 3 \times (4096 - 2)

=3×4094 = 3 \times 4094

=12282 = 12282


Using our formula with n=10n = 10:

S10=3×(210+1−2)S_{10} = 3 \times (2^{10+1} - 2)

=3×(211−2) = 3 \times (2^{11} - 2)

=3×(2048−2) = 3 \times (2048 - 2)

=3×2046 = 3 \times 2046

=6138 = 6138


Now we can find a11a_{11}:

a11=S11−S10a_{11} = S_{11} - S_{10}

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

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

=3×(4096−2−2048+2) = 3 \times (4096 - 2 - 2048 + 2)

=3×(2048) = 3 \times (2048)

=6144 = 6144

Answer: a11=6144a_{11} = 6144

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