# Solving a Recurrence Relation

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Solve the following recurrence relation:

a(n) = a(n-1) + 3(n-1), a(0) = 1

I know this should not be a difficult problem, but my main problem is in solving the problem when the coefficient of the a(n-1) term is 1. Also, when a summation is in the solution, I do not understand how to convert from a summation to a C(n,k) or a P(n,k). Could you please give a step by step explanation of the process involved with solving this type of problem? Thank you.

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##### Solution Summary

A recurrence relation is solved. The solution is detailed.

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a(1) = a(0) + 0 = 1

a(n) = a(n-1) + 3*(n-1).....

a(n) = a(n-2)+3*(n-2)+ 3*(n-1).....

a(n)= a(n-3) + 3*(n-3) + 3*(n-2) + 3*(n-1).....

and so on till u get a(1)...

So u will have:

a(n) = a(1) + 3*(1+2+3+4........+n-1)

Now sum of first k numbers is k(k+1)/2

with k=n-1 in this case.

so a(n) = 1 + [3*n(n-1)/2]

THIS IS THE SIMPLEST KIND OF RECURRENCE ...

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