# Find k_{513}

#### anemone

##### MHB POTW Director
Staff member
Let $k_1,k_2,\cdots$ be a sequence defined by $k_1=1$ and for $n \ge 1$, $k_{n+1}=\sqrt{k_n^2-2k_n+3}+1$. Find $k_{513}$.

##### Well-known member
Let $k_1,k_2,\cdots$ be a sequence defined by $k_1=1$ and for $n \ge 1$, $k_{n+1}=\sqrt{k_n^2-2k_n+3}+1$. Find $k_{513}$.
32

solved as

Take 1 to left and square both sides

(K(n+1) – 1)^2 = k(n)^2 – 2k(n) + 3 = (k(n)-1)^2 + 2
Or (K(n+1) – 1)^2 = (k(n)-1)^2 + 2
(k(2)-1)^2 = (k(1)-1)^2 + 2
(k(3)-1)^2 = (k(2)-1)^2 + 2 = = (k(1)-1)^2 + 2 * 2
Proceeding we see hat = (k(n)-1)^2 = = (k(1)-1)^2 + 2(n-1)
One can prove it by induction
Put n = 513 to get k(513) = (k(1)-1)^2 + 2 * 512 = 1024
K(513) = 32 as it has to be positive

#### anemone

##### MHB POTW Director
Staff member
32

solved as

Take 1 to left and square both sides

(K(n+1) – 1)^2 = k(n)^2 – 2k(n) + 3 = (k(n)-1)^2 + 2
Or (K(n+1) – 1)^2 = (k(n)-1)^2 + 2
(k(2)-1)^2 = (k(1)-1)^2 + 2
(k(3)-1)^2 = (k(2)-1)^2 + 2 = = (k(1)-1)^2 + 2 * 2
Proceeding we see hat = (k(n)-1)^2 = = (k(1)-1)^2 + 2(n-1)
One can prove it by induction
Put n = 513 to get k(513) = (k(1)-1)^2 + 2 * 512 = 1024
K(513) = 32 as it has to be positive

#### MarkFL

Staff member
Here is my solution:

We are given the recursive algorithm:

$$\displaystyle k_{n+1}=\sqrt{k_n^2-2k_n+3}+1$$ where $$\displaystyle k_1=1$$

If we subtract 1 from both sides and square, we obtain:

$$\displaystyle \left(k_{n+1}-1 \right)^2=\left(k_{n}-1 \right)^2+2$$

If we define:

$$\displaystyle U_n=\left(k_{n}-1 \right)^2$$

we then obtain the linear difference equation:

$$\displaystyle U_{n+1}-U_{n}=2$$ where $$\displaystyle U_1=0$$

The homogeneous solution is:

$$\displaystyle h_n=c_1$$

and the particular solution is:

$$\displaystyle p_n=c_2n$$

Substituting the particular solution into the difference equation, we find:

$$\displaystyle c_2(n+1)-c_2n=2\implies c_2=2$$

Thus, the general solution is:

$$\displaystyle U_n=c_1+2n$$

We may now use the initial value to determine the parameter $c_1$:

$$\displaystyle U_1=c_1+2=0\implies c_1=-2$$

And so the solution satisfying the given conditions is:

$$\displaystyle U_n=-2+2n=2(n-1)$$

Hence, we find:

$$\displaystyle U_{513}=2(513-1)=1024$$

Thus:

$$\displaystyle k_{513}=\sqrt{U_{513}}+1=33$$

#### anemone

##### MHB POTW Director
Staff member
Here is my solution:

We are given the recursive algorithm:

$$\displaystyle k_{n+1}=\sqrt{k_n^2-2k_n+3}+1$$ where $$\displaystyle k_1=1$$

If we subtract 1 from both sides and square, we obtain:

$$\displaystyle \left(k_{n+1}-1 \right)^2=\left(k_{n}-1 \right)^2+2$$

If we define:

$$\displaystyle U_n=\left(k_{n}-1 \right)^2$$

we then obtain the linear difference equation:

$$\displaystyle U_{n+1}-U_{n}=2$$ where $$\displaystyle U_1=0$$

The homogeneous solution is:

$$\displaystyle h_n=c_1$$

and the particular solution is:

$$\displaystyle p_n=c_2n$$

Substituting the particular solution into the difference equation, we find:

$$\displaystyle c_2(n+1)-c_2n=2\implies c_2=2$$

Thus, the general solution is:

$$\displaystyle U_n=c_1+2n$$

We may now use the initial value to determine the parameter $c_1$:

$$\displaystyle U_1=c_1+2=0\implies c_1=-2$$

And so the solution satisfying the given conditions is:

$$\displaystyle U_n=-2+2n=2(n-1)$$

Hence, we find:

$$\displaystyle U_{513}=2(513-1)=1024$$

Thus:

$$\displaystyle k_{513}=\sqrt{U_{513}}+1=33$$
Bravo, MarkFL and thanks for participating!

##### Well-known member
MarkFL and anemone
thanks

my solution
Proceeding we see hat = (k(n)-1)^2 = = (k(1)-1)^2 + 2(n-1)
I had done correct till the above

I forgot to take 1 to the right
(k(513) - 1)^2 = 1024

or k(513) = 33

so approach was right but not taking 1 to the right was an oversight

Last edited:

#### anemone

##### MHB POTW Director
Staff member
MarkFL and anemone
thanks

my solution
Proceeding we see hat = (k(n)-1)^2 = = (k(1)-1)^2 + 2(n-1)
I had done correct till the above

I forgot to take 1 to the right
k(513) - 1 = 1024

or k(513) = 33

so approach was right but not taking 1 to the right was an oversight
I am sorry kaliprasad...I checked your approach but couldn't locate the mistake (the honest kind, of course) and now everything seems perfect about your solution! Well done, kaliprasad!