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If 
a_(1)=3 and 
a_(n+1)=(a_(n))^(2)-1 then find the value of 
a_(3).
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 a3 a_{3} .\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 a3 a_{3} .\newlineAnswer:
  1. Given terms and formula: We are given the first term of the sequence, a1=3a_{1} = 3, and the recursive formula for the sequence, an+1=(an)21a_{n+1} = (a_{n})^2 - 1. To find a3a_{3}, we first need to find a2a_{2} using the given formula.
  2. Find a2a_{2}: Using the recursive formula an+1=(an)21a_{n+1} = (a_{n})^2 - 1, we substitute n=1n = 1 to find a2a_{2}:
    a2=(a1)21a_{2} = (a_{1})^2 - 1
    a2=(3)21a_{2} = (3)^2 - 1
    a2=91a_{2} = 9 - 1
    a2=8a_{2} = 8
  3. Find a3a_{3}: Now that we have a2a_{2}, we can use it to find a3a_{3} using the same recursive formula:\newlinea3=(a2)21a_{3} = (a_{2})^2 - 1\newlinea3=(8)21a_{3} = (8)^2 - 1\newlinea3=641a_{3} = 64 - 1\newlinea3=63a_{3} = 63

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