Find the last five digits of T_2014

[anemone]

Let $T_n$ be a sequence such that $T_1=1$ and $T_{n+1}=\lfloor T_n+\sqrt{T_n}+0.5 \rfloor$.

Find the last five digits of $T_{2014}$.

 

[chisigma]

Let $T_n$ be a sequence such that $T_1=1$ and $T_{n+1}=\lfloor T_n+\sqrt{T_n}+0.5 \rfloor$.

Find the last five digits of $T_{2014}$.

[sp]With some comfortable steps You find that is...

$\displaystyle T_{2\ n + 1} = n^{2} + n + 1$

$\displaystyle T_{2\ n } = T_{2\ n + 1} - n = n^{2} + 1$

...so that is $\displaystyle T_{2014} = 1007^{2} + 1 = 1014050$...[/sp]

Kind regards$\chi$ $\sigma$

 

[anemone]

[sp]With some comfortable steps You find that is...

$\displaystyle T_{2\ n + 1} = n^{2} + n + 1$

$\displaystyle T_{2\ n } = T_{2\ n + 1} - n = n^{2} + 1$

...so that is $\displaystyle T_{2014} = 1007^{2} + 1 = 1014050$...[/sp]

Kind regards$\chi$ $\sigma$

Thanks chisigma for participating and the answer you have is correct! Well done!:)

I am thinking it would be best if you can show us how you arrived at the equation $\displaystyle T_{2\ n + 1} = n^{2} + n + 1$...(Sun)

 

[chisigma]

Thanks chisigma for participating and the answer you have is correct! Well done!:)

I am thinking it would be best if you can show us how you arrived at the equation $\displaystyle T_{2\ n + 1} = n^{2} + n + 1$...(Sun)

Spoiler

Let suppose at first don't to consider the floor function and the constant .5 and also suppose that the sequence starts with the index n=0, so that we work on the difference equation...

$\displaystyle \Delta{n}= T_{n+1} - T_{n} = \sqrt{T_{n}}\ (1)$

The (1) can be approximated with the ODE...

$\displaystyle y^{\ '} = \sqrt{y}\ (2)$

...the solution of which is...

$\displaystyle y = \frac{x^{2} + 2\ c\ x + c^{2}}{4}\ (3)$Now setting x=2 n and with a proper choise of c we arive to write...

$T_{2\ n} \sim n^{2} + n + 1\ (4)$

Of corse I don't say that all that is 'rigorous' at cent per cent...

Kind regards

$\chi$ $\sigma$

 

[Albert1]

Let $T_n$ be a sequence such that $T_1=1$ and $T_{n+1}=\lfloor T_n+\sqrt{T_n}+0.5 \rfloor$.

Find the last five digits of $T_{2014}$.

Spoiler

$T_1=1$
$T_2=T_{2\times 1}=\lfloor 1+\sqrt1+0.5 \rfloor=2=1^2+1$.
$T_3=T_{2\times 1+1}=\lfloor 2+\sqrt2+0.5 \rfloor=3=1^2+1+1$.
$T_4=T_{2\times 2}=\lfloor 3+\sqrt 3+0.5 \rfloor=5=2^2+1$.
$T_5=T_{2\times 2+1}=\lfloor 5+\sqrt5+0.5 \rfloor=7=2^2+2+1$.
$T_6=T_{2\times 3}=\lfloor 7+\sqrt7+0.5 \rfloor=10=3^2+1$.
$T_7=T_{2\times 3+1}=\lfloor 10+\sqrt{10}+0.5 \rfloor=13=3^2+3+1$.
from above we may suggest:
$T_{2n}=n^2+1$.---(1)
$T_{2n+1}=n^2+n+1$----(2).
(here n=1,2,3,4,5,-------)
$\therefore T_{2014}=1007^2+1=1014049+1=1014050$
the proof of(1) and (2)we can use the method of induction