Using Binomial Theorem

Dec 2018
16
5
India
If n is a positive integer, prove that the integral part of [FONT=&quot]
[/FONT] is an even integer.
 
Dec 2018
16
5
India
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.
 
Nov 2013
6,611
2,962
California
I would use induction as the overall strategy.

In the course of proving $P_n \Rightarrow P_{n+1}$ you're going to have to apply the binomial theorem.

See where that leads you.
 
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Jun 2013
1,075
562
Lebanon
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

$\left\lfloor x_n\right\rfloor =2a_n$
 
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topsquark

Forum Staff
Jan 2006
11,509
3,414
Wellsville, NY
Sorry for the first post.It is very difficult to write equations and symbols in this forum.
We have a LaTeX system here. It's pretty easy to learn the basics, like what you were trying to post.

-Dan
 
Dec 2018
16
5
India
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.
 
Jun 2013
1,075
562
Lebanon
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.
\(\displaystyle I + f + f_1 = 2k\)

Here $k$ is not an integer

Also,

$f + f_1$ is not an integer and it does not equal $1$
 
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