Question

5, 11, 21, 43, 85, ?

• A. 185
• B. 170
• C. 171
• D. 181

Hint:

Look for alternating \(\pm 1\) after a common multiplier.

Answer 1

The Correct Option is C

The pattern doubles and alternately adds/subtracts 1: \[ 5\!\times\!2{+}1=11,\quad 11\!\times\!2{-}1=21,\quad 21\!\times\!2{+}1=43,\quad 43\!\times\!2{-}1=85. \] Next: \[ 85\!\times\!2{+}1=171. \] \[ \boxed{171} \]

Answer 2

Instead of doubling each term and alternately adding or subtracting \(1\), notice what happens when consecutive terms are added together.

\[ 5+11=16=2^4,\qquad 11+21=32=2^5,\qquad 21+43=64=2^6,\qquad 43+85=128=2^7. \] Each consecutive pair sums to a power of \(2\) that doubles every time. So the next pair should sum to \(2^8=256\), meaning the missing term satisfies \(85+x=256\), giving \(x=256-85=171\).

1. Option 185: \(85+185=270\), which is not \(256\), so this fails the pairwise-sum pattern.
2. Option 170: \(85+170=255\), just one short of \(256\), so this doesn't fit either.
3. Option 171: \(85+171=256=2^8\), exactly continuing the doubling pattern of sums.
4. Option 181: \(85+181=266\), overshooting \(256\), so this is also inconsistent.

The pairwise-sum-doubles-as-a-power-of-2 pattern pins the next term at \(171\).

Hence, the correct answer is 171.