Evaluate :

Solution:

To find: Value of $(\sqrt{3}+1)^{5}-(\sqrt{3}-1)^{5}$

Formula used: (I) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C{n} b^{n}$

$(a+1)^{5}={ }^{5} C_{0} a^{5}+{ }^{5} C_{1} a^{5-1} 1+{ }^{5} C_{2} a^{5-2} 1^{2}+{ }^{5} C_{3} a^{5-3} 1^{3}+{ }^{5} C_{4} a^{5-4} 1^{4}+{ }^{5} C_{5} 1^{5}$

$\Rightarrow{ }^{5} \mathrm{C}_{0} \mathrm{a}^{5}+{ }^{5} \mathrm{C}_{1} \mathrm{a}^{4}+{ }^{5} \mathrm{C}_{2} \mathrm{a}^{3}+{ }^{5} \mathrm{C}_{3} \mathrm{a}^{2}+{ }^{5} \mathrm{C}_{4} \mathrm{a}+{ }^{5} \mathrm{C}_{5} \ldots$ (i)

$(a-1)^{5}$

$=\left[{ }^{5} C_{0} a^{5}\right]+\left[{ }^{5} C_{1} a^{5-1}(-1)^{1}\right]+\left[{ }^{5} C_{2} a^{5-2}(-1)^{2}\right]+\left[{ }^{5} C_{3} a^{5-3}(-1)^{3}\right]+$

$\left[{ }^{5} C_{4} a^{5-4}(-1)^{4}\right]+\left[{ }^{5} C_{5}(-1)^{5}\right]$

$\Rightarrow{ }^{5} C_{0} a^{5}-{ }^{5} C_{1} a^{4}+{ }^{5} C_{2} a^{3}-{ }^{5} C_{3} a^{2}+{ }^{5} C_{4} a-{ }^{5} C_{5} \ldots$ (ii)

Subtracting (ii) from (i)

$(a+1)^{5}-(a-1)^{5}=\left[{ }^{5} C_{0} a^{5}+{ }^{5} C_{1} a^{4}+{ }^{5} C_{2} a^{3}+{ }^{5} C_{3} a^{2}+{ }^{5} C_{4} a+{ }^{5} C_{5}\right]-$

$\left[{ }^{5} C_{0} a^{5}-{ }^{5} C_{1} a^{4}+{ }^{5} C_{2} a^{3}-{ }^{5} C_{3} a^{2}+{ }^{5} C_{4} a-{ }^{5} C_{5}\right]$

$\Rightarrow 2\left[{ }^{5} \mathrm{C}_{1} \mathrm{a}^{4}+{ }^{5} \mathrm{C}_{3} \mathrm{a}^{2}+{ }^{5} \mathrm{C}_{5}\right]$

$\Rightarrow 2^{\left[\left(\frac{5 !}{1 !(5-1) !} a^{4}\right)+\left(\frac{5 !}{3 !(5-3) !} a^{2}\right)+\left(\frac{5 !}{5 !(5-5) !}\right)\right]}$

$\Rightarrow 2\left[(5) a^{4}+(10) a^{2}+(1)\right]$

$\Rightarrow 2\left[5 a^{4}+10 a^{2}+1\right]=(a+1)^{5}-(a-1)^{5}$

Putting the value of $a=\sqrt{3}$ in the above equation

$(\sqrt{3}+1)^{5}-(\sqrt{3}-1)^{5}=2\left[5(\sqrt{3})_{4}+10(\sqrt{3})_{2}+1\right]$

$\Rightarrow 2[(5)(9)+(10)(3)+1]$

$\Rightarrow 2[45+30+1]$

$\Rightarrow 152$

Ans) 152