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

If a1=4 a_{1}=4 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=4 a_{1}=4 and an=(an1)2+n a_{n}=\left(a_{n-1}\right)^{2}+n then find the value of a4 a_{4} .\newlineAnswer:
  1. Find a2a_{2}: Given a1=4a_{1} = 4, we need to find a4a_{4} using the recursive formula an=(an1)2+na_{n} = (a_{n-1})^2 + n. We will start by finding a2a_{2}.\newlinea2=(a1)2+2a_{2} = (a_{1})^2 + 2\newline=(4)2+2\quad = (4)^2 + 2\newline=16+2\quad = 16 + 2\newline=18\quad = 18
  2. Find a3a_{3}: Now we will find a3a_{3} using a2a_{2}.\newlinea3=(a2)2+3=(18)2+3=324+3=327a_{3} = (a_{2})^2 + 3 = (18)^2 + 3 = 324 + 3 = 327
  3. Find a4a_{4}: Finally, we will find a4a_{4} using a3a_{3}.\newlinea4=(a3)2+4=(327)2+4=106929+4=106933a_{4} = (a_{3})^2 + 4 = (327)^2 + 4 = 106929 + 4 = 106933

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