Question

If A and B are coefficients of xn in the expansions of (1+x) ²ⁿ and (1+x) ²ⁿ⁻¹ respectively, then A/B is equal to___

a. 1
b. 2
c. ½
d. 1/n

Answer 1

The coefficient A of x^n in the expansion of (1+x)^(2n) and coefficient B of x^n in the expansion of (1+x)^(2n-1) are calculated using the Binomial Theorem. Simplifying A/B leads to the conclusion that A/B is equal to 2.

Step-by-step explanation:

Let's find the coefficient of xn in the expansion of (1+x)2n and (1+x)2n-1 using the Binomial Theorem.

The Binomial Theorem states that:

(a + b)n = an + n*an-1*b + n*(n-1)/2!*an-2*b2 + ...

For (1+x)2n, the coefficient of xn, which is A, is given by:

A = 2nCn

Similarly, for (1+x)2n-1, the coefficient of xn, which is B, is given by:

B = (2n-1)Cn

To find A/B, divide the two coefficients:

A/B = (2nCn) / ((2n-1)Cn)

This simplifies to:

A/B = (2n)! / (n!(2n-n)!) * (n! (2n-1-n)!) / ((2n-1)!)

A/B = (2n)! / (n!)2 * (n!)2 / (2n-1)!

A/B = (2n) / n = 2

Therefore, A/B is equal to 2.