If n is a positive integer, prove that the integral part of [IMG]file:///C:/Users/user/AppData/Local/Temp/msohtmlclip1/01/clip_image002.png[/IMG] is an even integer.
If n is a positive integer, prove that the integral part of (5√5+11)^(2n+1) is an even no.
I tried a lot on this ques but ans coming to be odd, which is incorrect.
I doubt my soln. would be of any help to u all.
Sorry for the first post.It is very difficult to write equations and symbols in this forum.
using the binomial theorem it is easy to see that there are integers $a_n$ and $b_n$ such that
$\displaystyle x_n=\left(5\sqrt{5}+11\right)^{2n+1}=a_n+b_n\sqrt{ 5}$
and
$\displaystyle y_n=\left(5\sqrt{5}-11\right)^{2n+1}=-a_n+b_n\sqrt{5}$
therefore
$x_n=2a_n+y_n$ with $0<y_n<1$
so
$\left\lfloor x_n\right\rfloor =2a_n$
What is wrong in my solution:
〖(5√5+11)〗^(2n+1)=I + f ; where I is the integral part and f is a fraction;
〖(5√5-11)〗^(2n+1)=f1; where f1 is a pure fraction .
0<f<1
0<f1<1
On adding:
I+f+f1=2(2n+1C0(〖(5√5)^2n+1 ).......)
,i.e. I+f+f1=2k where k is a constant
0<f+f1<2
.i.e. f+f1=1
I=2k-1 ; which is a odd no.