Let a_{1}, a_{2}, ….., be a sequence defined by a_{1} = 1 for n>=1, an+1 = √(a_{n}^2 – 2a_{n} + 3) + 1. Find a_{513}.

 Let a₁ + a₂, ….., be a sequence defined by a₁ = 1 for n>=1, a_n+1 = √(a_n² – 2a_n + 3) + 1. Find a_513.

_Ans. 

First simplify: √(a_n² – 2a_n + 3) + 1

√(a_n² – 2a_n + 1 + 2) + 1

√((a_n – 1)² + 2) + 1

Lets first find out a₂:

Put n = 1

a₁₊₁ = √(a₁² – 2a₁ + 3) + 1

a₂  = √((1-1)² + 2) + 1

a₂ = √2 + 1

Likewise put n = 2

a₂₊₁ = √((a₂ – 1)² + 2) + 1

a₃ = √((√2 + 1 – 1)² + 2) + 1

a₃ = √(4) + 1

a₃ = 3

a₃₊₁ = √((a₃ – 1)² + 2) + 1

a₄ = √((3 – 1)² + 2) + 1

a₄ = √6 + 1

a_{4 + 1} = √((a₄ – 1)² + 2) + 1

a₅ = √((√6 + 1 – 1)² + 2) + 1

a₅ = √8 + 1

Now, we can notice that as we put the values of n

we get the result in the surd in the form of:

 √(2*n) + 1

To get a₅₁₃, we have to put n = 512

So, the result directly come out to be 

                                                           = √(2*512) + 1

                                                           = √1024 + 1

                                                           = 32 + 1

                                                           = 33

So, a₅₁₃ = 33