find all solutions of the recurrence relation
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 means a0, the first term, for example. The second term is a1 or a.
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.
More... Read More: ...