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1. |
If Sn=4Sn-1+12n, where S0=6 and S1=7, find the solution for the recurrence relation. |
A. | a<sub>n</sub>=7(2<sup>n</sup>) 29/6n6<sup>n</sup> |
B. | a<sub>n</sub>=6(6<sup>n</sup>)+6/7n6<sup>n</sup> |
C. | a<sub>n</sub>=6(3<sup>n+1</sup>) 5n |
D. | a<sub>n</sub>=nn 2/6n6<sup>n</sup> |
Answer» C. a<sub>n</sub>=6(3<sup>n+1</sup>) 5n | |