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Most Important Selected Qs for JEE AdvancedMathematicsBinomial Theorem

If (4+ 15 )^ n = I + f , where n is an odd natural number, l is an integer and 0 f 1 , then

Options

  1. AI is natural number
  2. BI is an even integer
  3. C(I+f)(1-f)=1
  4. D1-f=(4- 15 )^n

Correct answer

D. 1-f=(4- 15 )^n

Step-by-step solution

I + f =(4+ 15 )^n Let g =(4- 15 )^ n , then 0 g 1 aligned & I + f = ^ n C ₀ 4^n+ ^n C ₁ 4^ n -1 15 + ^n C ₂ 4^ n -2 15+ ^n C₃ 4^ n -3 ( 15 )^3+ . & g = ^ n C ₀ 4^n- ^n C₁ 4^ n -1 15 + ^n C₂ 4^ n -2 15- ^n C₃ 4^ n-3 ( 15 )^3+ . . aligned I + f + g =2 ( ^ n C ₀ 4^ n + ^ n C ₂ 4^ n -2 15+ .. )= even integer aligned & 0 f + g 2 & f + g =1 aligned 1- f = g thus I is an odd integer 1- f = g =(4- 15 )^ n (I+f)(1-f)=(I+f) g=1

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