Question: Given the recurrence relation T (n) =T(n 1) + 4n2 for n > 1, where T (0) = 8, what value of n will make

Given the recurrence relation T (n) =T(n 1) + 4n2 for

Given the recurrence relation T (n) =T(n 1) + 4n2 for n > 1, where T (0) = 8, what value of n will make the value of the recurrence relation 372? n(n+1) n(n+1)(2n+1) You may assume that I = 1=n+1, X=1 i= and =1 2 2 6 o 3 o O 8 O 7 5 9 6

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