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\(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\) is equal to1. 2522. 1523. 524. 765. 176 |
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Answer» Correct Answer - Option 2 : 152 Concept: Binomial expansion for \(\rm\left ( x+a \right )^{n}-\left ( x-a \right )^{n}\) is given by \(\rm\left ( x+a \right )^{n}-\left ( x-a \right )^{n}\) = 2{nC1xn-1a1 + nC3xn-3a3 + nC5xn-5a5 +....} \(\rm\left ( x+a \right )^{n}-\left ( x-a \right )^{n}\) = 2{sum of terms at even places} Calculation: For this expression \(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\) assume \(\rm \sqrt{3} =a\) \(\rm(a+1)^5-(a-1)^5\) = 2{5C1 a5-1 11 + 5C3 a5-3 13 + 5C5 a5-5 15} \(\rm(a+1)^5-(a-1)^5\) = 2{5C1a4 + 5C3a2+ 5C5} \(\rm(a+1)^5-(a-1)^5\) = 2{5a4 + 10a2+ 1} Put \(\rm a =\sqrt{3}\) \(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\) = 2{5(32) + 10(3)+ 1} \(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\) = 2{5.32 + 10.3 + 1} \(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\) = 2{45 + 30 +1} \(\rm(\sqrt{3}+1)^5-(\sqrt{3}-1)^5\) = 152 |
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