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Define P(n) = sum_(r=0)ⁿ [(-1)^r / (r+1)] * C(n,r). Which of the following are correct? (A) |P(10)| is the harmonic mean of |P(9)| and |P(11)|. (B) sum_(r=5)¹⁰ P(r) * P(r-1) = 6/55. (C) |P(10)| is the arithmetic mean of |P(9)| and |P(11)|. (D) sum_(r=5)¹⁰ P(r) * P(r-1) = 6/55.

1. (A) |P(10)| is the harmonic mean of |P(9)| and |P(11)|.
2. (B) sum from r=5 to 10 of P(r)*P(r-1) = 6/55.
3. (C) |P(10)| is the arithmetic mean of |P(9)| and |P(11)|.
4. (D) sum from r=5 to 10 of P(r)*P(r-1) = 6/55.

Correct answer: (A) |P(10)| is the harmonic mean of |P(9)| and |P(11)|.

Solution

Using the integral representation, P(n) = 1/(n+1). So P(9)=1/10, P(10)=1/11, P(11)=1/12. The harmonic mean of 1/10 and 1/12 equals 2/(10+12)=1/11=P(10), confirming A. The sum in B and D telescopes: sum_(r=5)¹⁰ 1/((r+1)r) = 1/5 - 1/11 = 6/55, confirming both B and D. C is false since the AM of 1/10 and 1/12 is not 1/11.

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