please any solve this sum from my engineering sem 3  subject discrete mathematics

find all solutions of the recurrence relation

an = 5an-1  +  6an-2  + 7n

 

 

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1 Answer

n=2: a2=5a1+6a0+14; n=3: a3=5a2+6a1+21; etc.

We do not have a0 or a1 to calculate a2 and further terms.
 

Assume the following:

"note for  5a(n-1)  in this (n-1) is to the base   same for 6a(n-2)  in this n-2 is to the base  and for 7n  in this n is to the power " which needs to be interpreted. So does this mean:

a[n]=5a[n-2]+6a[n-1]+7^n? The square brackets are meant to represent nth and other terms: a[0] means a0, the first term, for example. The second term is a1 or a[1].

If it does then: a2=5a0+6a1+7^2=5a0+6a1+49.

If there are no terms before a0, a0=1, a1=6+7=13; a2=5+78+49=132; a3=65+792+343=1200; etc.

The expression for a[n] can be split into 3 parts: 5a[n-2], 6a[n-1], 7^n.

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