If 8=3+1/4(3+p)+1/4²(3+2 p)+1/4³(3+3 p)+… … ∞, then the value of p is ____________.

Solution

Maths sir · 4 steps

Split every bracket into a 3-part and a p-part: one is a plain GP, the other a standard AGP.

1. Split the series. The general term is 1/4ⁿ(3 + np) for n = 0, 1, 2, …. Break it into two separate sums, one carrying the 3s and one carrying the ps.

   8 = 3Σₙ₌₀^∞1/4ⁿ + pΣₙ₌₁^∞n/4ⁿ

2. The GP part. Σₙ₌₀^∞1/4ⁿ = 1/(1 - 1/4) = 4/3, so the 3s add up to 3 × 4/3 = 4.
3. The AGP part. Use Σₙ₌₁^∞ n xⁿ = x/(1-x)² with x = 1/4. This gives (1/4)/(9/16) = 4/9, so the ps add up to 4p/9.
4. So. 8 = 4 + 4p/9 gives 4p/9 = 4, so p = 9.

The common slip: 11.25. Counting the 3s as only the leading 3, and forgetting the 3 inside every bracket, gives 8 = 3 + 4p/9 and so p = 11.25.

The idea to keep. When each term is (constant + n times something) over a power, split it into a GP plus an AGP. Remember Σ n xⁿ = x/(1-x)² for |x| < 1.

Answer from the NTA final answer key. Solution by PrepWiser.