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

If a1=2 a_{1}=2 and an+1=(an)22 a_{n+1}=\left(a_{n}\right)^{2}-2 then find the value of a4 a_{4} .\newlineAnswer:

Full solution

Q. If a1=2 a_{1}=2 and an+1=(an)22 a_{n+1}=\left(a_{n}\right)^{2}-2 then find the value of a4 a_{4} .\newlineAnswer:
  1. Find a2a_{2}: Use the given recursive formula to find a2a_{2}. The recursive formula is an+1=(an)22a_{n+1} = (a_{n})^2 - 2. We know that a1=2a_{1} = 2. Calculate a2a_{2} using a1a_{1}: a2=(a1)22=(2)22=42a_{2} = (a_{1})^2 - 2 = (2)^2 - 2 = 4 - 2.
  2. Find a3a_{3}: Use the recursive formula to find a3a_{3}. Now that we have a2a_{2}, we can find a3a_{3} using the same formula: a3=(a2)22=(4)22=162a_{3} = (a_{2})^2 - 2 = (4)^2 - 2 = 16 - 2.
  3. Find a4a_{4}: Use the recursive formula to find a4a_{4}. With a3a_{3} found, we can calculate a4a_{4}: a4=(a3)22=(16)22=2562a_{4} = (a_{3})^2 - 2 = (16)^2 - 2 = 256 - 2.

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