\(A_ 1 = 0\) and \(A_ 2 = 1\)

\(A_ n = A_{n-1} - A_{n-2}\)

=> \(A_ 3 = A_ 2 - A_ 1 = 1 - 0 = 1\)

=> \(A_ 4 = A_ 3 - A_ 2 = 1 - 1 = 0\)

=> \(A_ 5 = A_ 4 - A_ 3 = 0 - 1 = -1\)

=> \(A_ 6 = A_ 5 - A_ 4 = -1 - 0 = -1\)

=> \(A_ 7 = A_ 6 - A_ 5 = -1 - (-1) = -1 + 1 = 0\)

The sequence repeats 0, 1, 1, 0, -1, -1, 0, 1, 1 .....

The sequence repeats for every 6 terms and the sum of these six terms is 0 + 1 + 1 + 0 + (-1) + (-1) = 0

With in 100, there are 16 such groups till 96th term and the sum of all terms till 96th term is 0

=> \(A_{97} = A_{96} - A_{95} = -1 - (-1) = 0\)

Similarly \(A_{98} = 1\), \(A_{99} = 1\), \(A_{100} = 0\)

Sum of first hundred terms is \(0 + A_{97} + A_{98} + A_{99} + A_{100} = 0 + 0 + 1 + 1 + 0 = 2\)

Hence

option B
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