Binomial Theorem 2 Question 2

2. The value of r for which

20Cr20C0+20Cr120C1+20Cr220C2+.+20C020Cr is maximum, is

(2019 Main, 11 Jan I)

(a) 15

(b) 10

(c) 11

(d) 20

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Answer:

Correct Answer: 2. (d)

Solution:

  1. We know that,

(1+x)20=20C0+20C1x+20C2x2++20Cr1xr1+20Crxr++20C20x20(1+x)20(1+x)20=(20C0+20C1x+20C2x2++20Cr1xr1+20Crxr++20c20x20)×(20C0+20C1x++20Cr1xr1+20Crxr++20C20x20)(1+x)40=(20C020Cr+20C120Cr120Cr20C0)xr+

On comparing the coefficient of xr of both sides, we get

20C020Cr+20C120Cr1++20Cr20C0=40Cr

The maximum value of 40Cr is possible only when r=20

[nCn/2 is maximum when n is even ]

Thus, required value of r is 20 .



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