Recurrences part 4

A sequence starts with a 0 = 0 a_0=0 and a 1 = 1 a_1=1 .

If the further terms are given by

a n = 7 a n 1 10 a n 2 + n \displaystyle a_n= 7a_{n-1}- 10a_{n-2}+ n

Then find the last 3 digits of a 20 a_{20} .

Details :- If you don't know how to get the solutions for recurrence relations, you may try to learn it here .

This note will prove helpful in the coming problems of this series.

First Problem , Second problem and Third Problem are easier than this one.


The answer is 730.

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