MCQMediumJEE Main 2025 · 28 January, Shift 1Sum of Series

Mathematics Question from JEE Main 2025 · 28 January, Shift 1

Let an\langle a_n \rangle be a sequence such that a0=0a_0 = 0, a1=12a_1 = \frac{1}{2}, and 2an+2=5an+13an2a_{n+2} = 5a_{n+1} - 3a_n. n=0,1,2,3....n= 0,1,2,3.... Then k=1100ak\sum_{k=1}^{100} a_k is equal to:

  • A

    3a99+1003a_{99} + 100

  • B

    3a991003a_{99} - 100

  • C

    3a100+1003a_{100} + 100

  • D

    3a1001003a_{100} - 100

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