InterviewSolution
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InterviewSolution
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Let `(1 + x)^(36) = a_(0) +a_(1) x + a_(2) x^(2) +...+ a_(36)x^(36)`. Then Statememt-1: `a_(0) +a_(3) +a_(6) +…+a_(36)= (2)/(3) (2^(36) +1)` Statement-2: `a_(0) + a_(2) +a_(4) +…+ a_(36) = 2^(35)`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - b We have, `a_(0) + a_(1) + a_(2) x^(2) + a_(3) + … + a_(36) x^(36) = (1 + x)^(36)` …(i) Putting x = 1 `omega, omega^(2)` successively, we get `(a_(0) + a_(3) + a_(6) +...+ a_(36) )+ (a_(1) + a_(4) + a_(7) +...+a_(34))+(a_(2) + a_(5) +...+a_(35)) = 2^(36)` ...(ii) `(a_(0) + a_(3) + a_(6) +...+ a_(36) )+omega (a_(1) + a_(4) + a_(7) +...+a_(34))+omega^(2)(a_(2) + a_(5) +...+a_(35)) = 1` ...(iii) `(a_(0) + a_(3) + a_(6) +...+ a_(36) )+omega^(2) (a_(1) + a_(4) + a_(7) +...+a_(34))+omega(a_(2) + a_(5) +...+a_(35)) = 1` `(a_(0) + a_(3) + a_(6) +...+ a_(36) )+omega^(2) (a_(1) + a_(4) + a_(7) +...+a_(34))+omega(a_(2) + a_(5) +...+a_(35)) = 1` ....(iv) Adding these three , we get `3(a_(0) + a_(3) + a_(6) +....+ a_(36)) = 2 (2^(35) +1)` `rArr a_(0) + a_(3) + a_(6) +....+a_(36) = (2)/(3) (2^(35) +1)` So, statement-1 is true. Putting x = -1 in (i) and adding it to (ii) , we get `a_(0) + a_(2) + a_(4) +...+ a_(36) = 2^(35)` So, statement-2 is true. But , statement-2 is not a correct explanation for satatement -1 . |
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