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If 
a_(1)=3 and 
a_(n+1)=(a_(n))^(2)-1 then find the value of 
a_(4).
Answer:

If a1=3 a_{1}=3 and an+1=(an)21 a_{n+1}=\left(a_{n}\right)^{2}-1 then find the value of a4 a_{4} .\newlineAnswer:

Full solution

Q. If a1=3 a_{1}=3 and an+1=(an)21 a_{n+1}=\left(a_{n}\right)^{2}-1 then find the value of a4 a_{4} .\newlineAnswer:
  1. Identify First Term: Identify the first term in the sequence.\newlineThe first term a1a_{1} is given as 33.
  2. Find Second Term: Use the recursive formula to find the second term a2a_{2}. The recursive formula is an+1=(an)21a_{n+1} = (a_{n})^2 - 1. Substitute n=1n=1 into the formula to find a2a_{2}. a2=(a1)21=(3)21=91=8a_{2} = (a_{1})^2 - 1 = (3)^2 - 1 = 9 - 1 = 8.
  3. Find Third Term: Use the recursive formula to find the third term a3a_{3}. Substitute n=2n=2 into the formula to find a3a_{3}. a3=(a2)21=(8)21=641=63a_{3} = (a_{2})^2 - 1 = (8)^2 - 1 = 64 - 1 = 63.
  4. Find Fourth Term: Use the recursive formula to find the fourth term a4a_{4}. Substitute n=3n=3 into the formula to find a4a_{4}. a4=(a3)21=(63)21=39691=3968a_{4} = (a_{3})^2 - 1 = (63)^2 - 1 = 3969 - 1 = 3968.

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