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If 
a_(1)=2,a_(2)=0 and 
a_(n)=3a_(n-1)-3a_(n-2) then find the value of 
a_(5).
Answer:

If a1=2,a2=0 a_{1}=2, a_{2}=0 and an=3an13an2 a_{n}=3 a_{n-1}-3 a_{n-2} then find the value of a5 a_{5} .\newlineAnswer:

Full solution

Q. If a1=2,a2=0 a_{1}=2, a_{2}=0 and an=3an13an2 a_{n}=3 a_{n-1}-3 a_{n-2} then find the value of a5 a_{5} .\newlineAnswer:
  1. Initialize Values: To find the value of a5a_{5}, we need to use the recursive formula an=3an13an2a_{n}=3a_{n-1}-3a_{n-2} and the initial values a1=2a_{1}=2 and a2=0a_{2}=0. We will calculate the values of a3a_{3}, a4a_{4}, and then a5a_{5} in sequence.
  2. Calculate a3a_{3}: First, let's find a3a_{3} using the recursive formula. We have a3=3a23a1a_{3}=3a_{2}-3a_{1}. Substituting the known values, we get a3=3×03×2a_{3}=3\times 0-3\times 2.
  3. Calculate a4a_{4}: Calculating a3a_{3}, we get a3=06=6a_{3}=0-6=-6.
  4. Calculate a5a_{5}: Next, we find a4a_{4} using the recursive formula. We have a4=3a33a2a_{4}=3a_{3}-3a_{2}. Substituting the known values, we get $a_{\(4\)}=\(3\)*(\(-6\))\(-3\)*\(0\).
  5. Final Result: Calculating \(a_{4}\), we get \(a_{4}=-18-0=-18\).
  6. Final Result: Calculating \(a_{4}\), we get \(a_{4}=-18-0=-18\). Finally, we find \(a_{5}\) using the recursive formula. We have \(a_{5}=3a_{4}-3a_{3}\). Substituting the known values, we get \(a_{5}=3*(-18)-3*(-6)\).
  7. Final Result: Calculating \(a_{4}\), we get \(a_{4}=-18-0=-18\). Finally, we find \(a_{5}\) using the recursive formula. We have \(a_{5}=3a_{4}-3a_{3}\). Substituting the known values, we get \(a_{5}=3*(-18)-3*(-6)\). Calculating \(a_{5}\), we get \(a_{5}=-54+18=-36\).

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