Use the binomial theorem to find the exact value of $$\left( \sqrt{3}+1 \right)^{5}-\left( \sqrt{3}-1 \right)^{5}$$ Hence show, without using tables or calculators, that the value of $$\left( \sqrt{3}+1 \right)^{5}$$ lies between 152 and 153.

Question

trojanpoem · Answer

Number of terms = n+1 = 5+1 = 6

(1+sqrt(3))^5 + (1- sqrt(3))^5

(1+sqrt(3))^5 = 1 - sqrt(3) * 5 + sqrt(3) * 5 * 4 / 2! - sqrt(3) * 5 * 4 * 3 /3! + sqrt(3) * 5 * 4 * 3  * 2/3! - sqrt(3) * 5 * 4 * 3 * 2 * 1/4! 

(1- sqrt(3))^5= 1 + sqrt(3) * 5 + sqrt(3) * 5 * 4 / 2! + sqrt(3) * 5 * 4 * 3 /3! + sqrt(3) * 5 * 4 * 3  * 2/3! +  sqrt(3) * 5 * 4 * 3 * 2 * 1/4!

So we will get 2 * ( the odd terms) , we could have induced that at first, as It's common in binomials.


2 ( 1 + sqrt(3)*5*4/2! + sqrt(3) * 5 * 4 * 3 * 2 * 1/(3!)  )  << 6/2 = 3 terms

(2 + 20sqrt(3) + 40sqrt(3) )

rayaz · Answer

Shouldn't your final answer be $$2+20\left( \sqrt{3} \right)^{2}+10\left( \sqrt{3} \right)^{4}$$

ganeshie8 · Answer

Hey!

rayaz · Answer

Hi

ganeshie8 · Answer

$$\left( \sqrt{3}+1 \right)^{5}-\left( \sqrt{3}-1 \right)^{5} \~\ 
= \sum\limits_{k=0}^5 \dbinom{5}{k}(\sqrt{3})^k (1)^{5-k} - \sum\limits_{k=0}^5 \dbinom{5}{k}(\sqrt{3})^k (-1)^{5-k}\~\

= \sum\limits_{k=0}^5 \dbinom{5}{k}(\sqrt{3})^k [(1)^{5-k}-(-1)^{5-k}] \~\
= \sum\limits_{k=0}^5 \dbinom{5}{k}(\sqrt{3})^k [1-(-1)^{5-k}] \~\
$$

Clearly, the terms cancel out when $k$ is odd. Therefore the sum is :
$$2\left[\dbinom{5}{0}(\sqrt{3})^0+\dbinom{5}{2}(\sqrt{3})^2+\dbinom{5}{4}(\sqrt{3})^4\right]$$

simplify

rayaz · Answer

Thanks for the explanation, can you help with the second part of the question. I don't know how to approach it. This is ow I've been trying to solve it 
$$\left( \sqrt{3}+1 \right)^{5}\ =152+\left( \sqrt{3}-1 \right)^{5}$$
After this I'm stuck.

ganeshie8 · Answer

Very good!

ganeshie8 · Answer

for second part, just notice this : 

$\sqrt{3}-1$ is between $0$ and $1$

rayaz · Answer

Oh! Well that was clumsy of me to not notice. Thanks.

ganeshie8 · Answer

$$0\lt \sqrt{3}-1\lt 1$$
$$0\lt (\sqrt{3}-1)^5\lt 1$$
$$152+0\lt 152+(\sqrt{3}-1)^5\lt 152+ 1$$

ganeshie8 · Answer

There is another nice way to work this problem with out using binomial theorem

ganeshie8 · Answer

If you're interested, I can give you one or two hints :)

rayaz · Answer

Go ahead

ganeshie8 · Answer

Hint 1 : 
Notice that $1\pm \sqrt{3}$ are the roots of quadratic $x^2-2x-2=0$

ganeshie8 · Answer

Let $r = 1+\sqrt{3}$ and $s=1-\sqrt{3}$

Since these are the roots of above quadratic, we have : 
$$s^2-2s-2=0\implies s^2 = 2s+2\tag{1}$$$$r^2-2r-2=0\implies r^2 = 2r+2\tag{2}$$

ganeshie8 · Answer

multiplying both sides by $s^{k-2}$ and $r^{k-2}$ gives

$$s^k = 2s^{k-1}+2s^{k-2}\tag{3}$$$$ r^k = 2r^{k-1}+2r^{k-2}\tag{4}$$

ganeshie8 · Answer

Adding (3) and (4) :

$$s^k+r^k = 2(s^{k-1}+r^{k-1})+2(s^{k-2}+r^{k-2})\tag{5}$$

ganeshie8 · Answer

we already know that $s+r = 2$

plugin $k=2$ in (5) :
$$s^2+r^2 \= 2(s^{2-1}+r^{2-1})+2(s^{2-2}+r^{2-2})\=2(s+r)+2(1+1)\=2(2)+2(2)\=8$$

ganeshie8 · Answer

plugin $k=3$ in (5) :
$$s^3+r^3 \= 2(s^{3-1}+r^{3-1})+2(s^{3-2}+r^{3-2})\=2(s^2+r^2)+2(s+r)\=2(8)+2(2)\=20$$

ganeshie8 · Answer

plugin $k=4$ in (5) :
$$s^4+r^4 \= 2(s^{4-1}+r^{4-1})+2(s^{4-2}+r^{4-2})\=2(s^3+r^3)+2(s^2+r^2)\=2(20)+2(8)\=56$$

ganeshie8 · Answer

plugin $k=5$ in (5) :
$$s^5+r^5 \= 2(s^{5-1}+r^{5-1})+2(s^{5-2}+r^{5-2})\=2(s^4+r^4)+2(s^3+r^3)\=2(56)+2(20)\=152$$

ganeshie8 · Answer

As you can see we can find sum of power of roots recursively using above trick

rayaz · Answer

I'm gonna try this the next time I see a similar question. Once again, thanks a bunch.

ganeshie8 · Answer

np :) do feel free to tag me in all your future questions !

rayaz · Answer

Ok