If (2+sqrt{3})^{49}+(sqrt{3}-2)^{49}=a+b sqrt | vedclass

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(B) Let $x = (2+\sqrt{3})^{49} + (\sqrt{3}-2)^{49}$.Since $(\sqrt{3}-2)^{49} = - (2-\sqrt{3})^{49}$,we have $x = (2+\sqrt{3})^{49} - (2-\sqrt{3})^{49}

Vedclass · Accepted Answer

$b 
eq 0, a=0$

Vedclass · Answer

(B) Let $x = (2+\sqrt{3})^{49} + (\sqrt{3}-2)^{49}$.Since $(\sqrt{3}-2)^{49} = - (2-\sqrt{3})^{49}$,we have $x = (2+\sqrt{3})^{49} - (2-\sqrt{3})^{49}$.Using the binomial expansion $(x+y)^n - (x-y)^n = 2 \sum_{k=0, k 	ext{ is odd}}^{n} \binom{n}{k} x^{n-k} y^k$,where $n=49, x=2, y=\sqrt{3}$.$x = 2 [ \binom{49}{1} 2^{48} (\sqrt{3})^1 + \binom{49}{3} 2^{46} (\sqrt{3})^3 + \dots + \binom{49}{49} (\sqrt{3})^{49} ]$.Each term in the expansion contains an odd power of $\sqrt{3}$,which results in a multiple of $\sqrt{3}$.Thus,$x = 0 + b\sqrt{3}$,where $b 
eq 0$ and $a = 0$.Therefore,the correct option is $b 
eq 0, a=0$.

If $(2+\sqrt{3})^{49}+(\sqrt{3}-2)^{49}=a+b \sqrt{3}$,where $a, b \in \mathbb{Q}$,then

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