Exams Joint Entrance Examination Mathematics Binomial Theorem Question 18 Question 18 of 347 Medium MCQ
For an integer n ≥ 2 n\ge 2 n ≥ 2 , if the arithmetic mean of all coefficients in the binomial expansion of ( x + y ) 2 n − 3 (x+y{)}^{2n-3} ( x + y ) 2 n − 3 is 16 , then the distance of the point P ( 2 n − 1 , n 2 − 4 n ) P(2n-1,{n}^{2}-4n) P ( 2 n − 1 , n 2 − 4 n ) from the line x + y = 8 x+y=8 x + y = 8 is
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= 2 n − 3 C 0 + 2 n − 3 C 1 + 2 n − 3 C 2 + ⋯ 2 n − 3 C 2 n − 3 2 n − 2 = 16 = 2 2 n − 3 = 16 ( 2 n − 2 ) = 2 2 n − 3 = 2 5 ( n − 1 ) ⇒ n = 5 ∴ P ( 9 , 5 ) d = ∣ 9 + 5 − 8 2 ∣ = 6 2 = 3 2 \begin{gathered} =\frac{{}^{2n-3}{C}_{0}+{}^{2n-3}{C}_{1}+{}^{2n-3}{C}_{2}+\cdots {}^{2n-3}{C}_{2n-3}}{2n-2}=16 \\ ={2}^{2n-3}=16(2n-2) \\ ={2}^{2n-3}={2}^{5}(n-1) \\ \Rightarrow n=5 \\ \therefore \, P(9,5) \\ d=|\frac{9+5-8}{\sqrt{2}}|=\frac{6}{\sqrt{2}}=3\sqrt{2} \end{gathered} = 2 n − 2 2 n − 3 C 0 + 2 n − 3 C 1 + 2 n − 3 C 2 + ⋯ 2 n − 3 C 2 n − 3 = 16 = 2 2 n − 3 = 16 ( 2 n − 2 ) = 2 2 n − 3 = 2 5 ( n − 1 ) ⇒ n = 5 ∴ P ( 9 , 5 ) d = ∣ 2 9 + 5 − 8 ∣ = 2 6 = 3 2 Answer the question to unlock