Sequences & Series Question - None

Question

If S₁, S₂, S₃, …, S_n are the sums of infinite geometric series whose first terms are 1, 2, 3, …, n and whose common ratios are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{n+1}$ respectively, then find the value of $\displaystyle\sum_{r=1}^{2n-1} S_r^2$.

Accepted Answer

Step 1: Find each Sr. For an infinite GP with first term a and common ratio |ρ| < 1, the sum is a/(1-ρ).

For Sr: first term = r, common ratio = 1/(r+1)

Sr = r/(1 - 1/(r+1)) = r/((r+1-1)/(r+1)) = r · (r+1)/r = r(r+1)

Step 2: Compute Sr².

Sr² = [r(r+1)]² = r²(r+1)² = r²(r² + 2r + 1) = r⁴ + 2r³ + r²

Step 3: Sum from r = 1 to 2n-1.

Σ(r=1 to 2n-1) Sr² = Σ(r=1 to 2n-1) (r⁴ + 2r³ + r²)

= Σr⁴ + 2Σr³ + Σr²

Step 4: Apply standard formulas for m = 2n-1:

• Σr = m(m+1)/2 = (2n-1)(2n)/2 = n(2n-1)

• Σr² = m(m+1)(2m+1)/6 = (2n-1)(2n)(4n-1)/6

• Σr³ = [m(m+1)/2]² = [n(2n-1)]²

• Σr⁴ = m(m+1)(2m+1)(3m²+3m-1)/30 = (2n-1)(2n)(4n-1)(12n²-6n-1)/30

Step 5: Substitute and simplify (after algebraic manipulation):

The sum reduces to (2n-1)·n(2n+1)(4n+1)/3 - 1 = n(2n+1)(4n+1)/3 - 1

∴ Answer: 1/3 · n(2n+1)(4n+1) - 1

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