Previous Year Paper
CAT2005Previous Years Paper
If a_{1} = 1 and a_{n}_{+1} − 3a_{n} + 2 = 4n for every positive integer n, then a_{100} equals
A  3^{99} − 200 
B  3^{99} + 200

C  3^{100} − 200 
D  3^{100} + 200

a_{1}= 1, a_{n}_{+1} – 3a_{n} + 2 = 4n
A_{n}_{+1} = 3a_{n} + 4n – 2
When n = 2, then a^{2} = 3 + 4 – 2 = 5
When n = 3, then a_{3} = 3 × 5 + 4 × 2 – 2 = 21
From the options, we get an idea that a_{n} can be expressed in a combination of some power of 3 and some multiple of 100.
(a) 3^{99} – 200; tells us that a_{n} could be: 3^{n}^{}^{1}−2 × n; but it does not fit a_{1} or a_{2} or a_{3}
(b) 3^{99} + 200; tells us that a_{n} could be: 3^{n}^{}^{1}+ 2 × n; again not valid for a_{1}, a_{2} etc.
(c) 3^{100} + 200; tells 3^{n} + 2n: again not valid.
So, (c) is the correct answer.
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