A) 134 and 135
B) 135 and 136
C) 136 and 137
D) None of these
Correct Answer: B
Solution :
\[{{(2+\sqrt{2})}^{4}}={{(\sqrt{2})}^{4}}{{(\sqrt{2}+1)}^{4}}\] = \[4\,{{[}^{4}}{{C}_{0}}{{+}^{4}}{{C}_{1}}(\sqrt{2}){{+}^{4}}{{C}_{2}}{{(\sqrt{2})}^{2}}{{+}^{4}}{{C}_{3}}{{(\sqrt{2})}^{3}}{{+}^{4}}{{C}_{4}}{{(\sqrt{2})}^{4}}]\] = \[4\,\left[ 1+4\sqrt{2}+\frac{4.3}{2}.2+\frac{4.3.2}{1.2.3}.2\sqrt{2}+4 \right]\] = \[4\,[1+4\sqrt{2}+12+8\sqrt{2}+4]\] = \[4\,[17+12\sqrt{2}]\] = \[4\,[17+(\tilde{-}17)]\] = \[4\,[\,34]=136\].
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