1. ## Using Binomial Theorem

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.

2. ## Re: Using Binomial Theorem

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.

3. ## Re: Using Binomial Theorem

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.

4. ## Re: Using Binomial Theorem

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

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

5. ## Re: Using Binomial Theorem

Originally Posted by mak29
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

6. ## Re: Using Binomial Theorem

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

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.

7. ## Re: Using Binomial Theorem

Originally Posted by mak29
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

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 $1$