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Most Important Selected Qs for JEE AdvancedMathematicsBinomial Theorem

Consider (1+x)^ 2 n + (1+2 x+x^2 )^n= _ r=0 ^ 2 n a_r x^r, n N . If _ r=0 ^ 2 n a_r=f(n) then:

Options

  1. A_ n=1 ^ 1 f(n) = 1 6
  2. B_ n=1 ^ 1 f(n) = 3 8
  3. Clargest value of p for which f(5) is divisible by 2^p is 11 .
  4. Dlargest value of p for which f(5) is divisible by 2^p is 9 .

Correct answer

C. largest value of p for which f(5) is divisible by 2^p is 11 .

Step-by-step solution

(1+x)^ 2 n + (1+2 x+x^2 )^n= _ r=0 ^ 2 n a_r x^r2(1+x)^ 2 n = _ r=0 ^ 2 n a_r x^r So, f(n)= _ r=0 ^ 2 n a_r=2^ 2 n+1 ...(1) So, array r _ n=1 ^ 1 f(n) = 1 2^3 + 1 2^5 + 1 2^7 + . . = 1 / 2^3 1- 1 4 = 1 / 8 3 / 4 = 1 6 array Largest value of p for which f(5) is divisible by 2^pf(5)=2¹¹ So, p=11

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