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The greatest integer less than or equal to `(sqrt2+1)^6` isA. 197B. 198C. 196D. 199 |
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Answer» Correct Answer - a Let `(2sqrt(2) +1)^(6) = I + F` where I `in N and 0 lt F lt 1`. Clearly , I is the greatest integer less then or equal to `(2 sqrt(2) +1)^(6)` Let G =` (sqrt(2) -1)^(6)` . Then , `I + F + G = (sqrt(2) + 1)^(6) + (sqrt(2) -1)^(6)` `rArr I + F + G = 2 {""^(6)C_(0) (sqrt(2))^(6) + ""^(6)C_(2) (sqrt(2))^(6-2) +....}` `rArr I + F + G ` = An even integer `rArr F + G = 1` Again, `I + F + G` `2={""^(6)C_(0)(sqrt(2))^(6) + ""^(6)C_(2) (sqrt(2))^(6-2) + ""^(6)C_(4) (sqrt(2))^(6-4) + ""^(6)C_(6) (sqrt(2))^(6-6)}` `rArr I + 1 = 2 (8 + 15 xx4 + 15 xx4 + 15 xx2 +1) [because F + G = 1 ]` `rArr I = 197` |
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