If a₁ = 1 and a_n+1 – 3a_n + 2 = 4n for every positive integer n, then a₁₀₀ equals (CAT 2005)

3⁹⁹ – 200

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3⁹⁹ + 200

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3¹⁰⁰ – 200

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3¹⁰⁰ + 200

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Solution

The correct option is C 3¹⁰⁰ – 200
Option (c)
Shortcut: Double Substitution
Write each answer option in terms of n (where n=100)
3^n-1 – 2n
3^n-1 + 2n
3ⁿ - 2n
3ⁿ + 2n
Now at n=1, we should get 1. Put n=1 in the answer options and eliminate those options where you do not get 1
-1 ≠1
2 ≠1
1
5≠1 thus, the answer is option (c)

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