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

If f(1)=1 f(1)=1 and f(n)=f(n1)2+n f(n)=f(n-1)^{2}+n then find the value of f(4) f(4) .\newlineAnswer:

Full solution

Q. If f(1)=1 f(1)=1 and f(n)=f(n1)2+n f(n)=f(n-1)^{2}+n then find the value of f(4) f(4) .\newlineAnswer:
  1. Find f(2)f(2): Given f(1)=1f(1) = 1, we need to find f(4)f(4) using the recursive formula f(n)=f(n1)2+nf(n) = f(n-1)^{2} + n. First, we find f(2)f(2). f(2)=f(21)2+2f(2) = f(2-1)^{2} + 2 f(2)=f(1)2+2f(2) = f(1)^{2} + 2 f(2)=12+2f(2) = 1^{2} + 2 f(2)=1+2f(2) = 1 + 2 f(2)=3f(2) = 3
  2. Find f(3)f(3): Next, we find f(3)f(3) using the value of f(2)f(2).
    f(3)=f(31)2+3f(3) = f(3-1)^{2} + 3
    f(3)=f(2)2+3f(3) = f(2)^{2} + 3
    f(3)=32+3f(3) = 3^{2} + 3
    f(3)=9+3f(3) = 9 + 3
    f(3)=12f(3) = 12
  3. Find f(4)f(4): Finally, we find f(4)f(4) using the value of f(3)f(3).
    f(4)=f(41)2+4f(4) = f(4-1)^{2} + 4
    f(4)=f(3)2+4f(4) = f(3)^{2} + 4
    f(4)=122+4f(4) = 12^{2} + 4
    f(4)=144+4f(4) = 144 + 4
    f(4)=148f(4) = 148

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