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

If a1=2 a_{1}=2 and an=(an1)2+n a_{n}=\left(a_{n-1}\right)^{2}+n then find the value of a4 a_{4} .\newlineAnswer:

Full solution

Q. If a1=2 a_{1}=2 and an=(an1)2+n a_{n}=\left(a_{n-1}\right)^{2}+n then find the value of a4 a_{4} .\newlineAnswer:
  1. Calculate a2a_{2}: Given the recursive formula an=(an1)2+na_{n}=(a_{n-1})^{2}+n and the initial condition a1=2a_{1}=2, we need to find the value of a4a_{4}. We will start by finding a2a_{2} using the given formula.\newlinea2=(a1)2+2a_{2} = (a_{1})^2 + 2\newline =(2)2+2= (2)^2 + 2\newline =4+2= 4 + 2\newline =6= 6
  2. Calculate a3a_{3}: Next, we will find a3a_{3} using the value of a2a_{2} we just calculated.\newlinea_{3} = (a_{2})^2 + 3\(\newline = (6)^2 + 3\newline = 36 + 3\newline = 39\)
  3. Calculate a4a_{4}: Finally, we will find a4a_{4} using the value of a3a_{3}.a4=(a3)2+4=(39)2+4=1521+4=1525a_{4} = (a_{3})^2 + 4 = (39)^2 + 4 = 1521 + 4 = 1525

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