## Discussion Forum

Que. | Consider the following recurrence relation. The value of T(m^2) for m ≥ 1 is |

a. | (m/6) (21m - 39) + 4 |

b. | (m/6) (4m^2 - 3m + 5) |

c. | (m/2) (m^2.5 - 11m + 20) - 5 |

d. | (m/6) (5m^3 - 34m^2 + 137m - 104) + (5/6) |

Answer:(m/6) (4m^2 - 3m + 5) |