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- Feb 14, 2012

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32

solved as

(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

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Thanks for participating32

solved as

(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

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\(\displaystyle k_{n+1}=\sqrt{k_n^2-2k_n+3}+1\) 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:

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

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- Feb 14, 2012

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Bravo,

\(\displaystyle k_{n+1}=\sqrt{k_n^2-2k_n+3}+1\) 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:

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

- Mar 31, 2013

- 1,322

MarkFL and anemoneThanks for participatingkaliprasad...but your answer isn't correct. I'm sorry.

thanks

my solution

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:

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I am sorryMarkFL 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