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gwd 26-29

Q29:

If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1
= 2xn – 1 for n ≥ 1, then x20 x19 =
A. 219
B. 220
C. 221
D. 220 - 1
E. 221 - 1
我怎么都算不出答案,谢谢各位大侠

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X20 - X19
= (2 * X19 - 1) - X19
= X19 - 1
= (2 * X18 - 1) - 1
= 2 * (2 * X17 - 1) - 2
= 2 * (2 * (2 * X16 - 1) - 1) - 2
= 2 * 2 * 2 * X16 - (2 * 2 - 2 - 1) - 1
= 2^18 * X1 - (2^17 - 2^16 - ... - 2^2 - 2^1 - 2^0) - 1
= 2^18 * 3 - sum of(2^k) where k goes from 0 to 17 (see note) - 1
= 2^18 * 3 - 2^0 * (2^18 - 1) / (2 - 1) - 1
= 2^18 * 3 - 1 * 2^18 + 1 - 1
= 2^18 * 2
= 2^19

Note: sum of (x^k) where k goes from y to z = x^y * (x^(z - y + 1) - 1) / (x - 1)

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