Some Special Sequences — Practice Quiz

n(n+1)(n+2)(n+3)/4

Step 1: tₙ = n³+3n²+2n = n(n²+3n+2) = n(n+1)(n+2). Step 2: Sₙ = Σn(n+1)(n+2). Step 3: This is a product of 3 consecutive integers. Use Σn(n+1)(n+2) = n(n+1)(n+2)(n+3)/4. Step 4: Alternatively, expand: Σ(n³+3n²+2n) = [n(n+1)/2]²+3·n(n+1)(2n+1)/6+2·n(n+1)/2. Step 5: After simplification, both methods give n(n+1)(n+2)(n+3)/4. This elegant formula applies to products of consecutive integers.

10/21

Step 1: The nth term tₙ = 1/[(2n-1)(2n+1)]. Use partial fractions: 1/[(2n-1)(2n+1)] = (1/2)[1/(2n-1) - 1/(2n+1)]. Step 2: Write out terms: t₁=(1/2)[1-1/3], t₂=(1/2)[1/3-1/5], ..., t₁₀=(1/2)[1/19-1/21]. Step 3: This is a telescoping series. Sum = (1/2)[1 - 1/(2×10+1)] = (1/2)[1 - 1/21]. Step 4: = (1/2)(20/21) = 10/21. Step 5: The key technique is partial fractions leading to telescoping cancellation.

n(n+1)(n+2)(3n+7)/12

Step 1: The nth term: tₙ = (n+1)(n+2)² = (n+1)(n²+4n+4) = n³+5n²+8n+4. Step 2: Sₙ = Σn³ + 5Σn² + 8Σn + 4Σ1. Step 3: = n²(n+1)²/4 + 5n(n+1)(2n+1)/6 + 8n(n+1)/2 + 4n. Step 4: Factor out n(n+1)/12: = n(n+1)[3n(n+1) + 10(2n+1) + 48]/12 + 4n = n(n+1)[3n²+3n+20n+10+48]/12 + 4n. Step 5: After careful algebra, this simplifies to n(n+1)(n+2)(3n+7)/12.

Σr³ is always equal to Σr for any n

Step 1: The identity states Σr³ = (Σr)². Step 2: Σr = n(n+1)/2, so Σr³ = [n(n+1)/2]² = n²(n+1)²/4. This confirms option A is valid. Step 3: For n=10: Σr³ = 55² = 3025 and (Σr)² = 55² = 3025, confirming option B. Step 4: For n=1: Σr³ = 1 = Σr = 1, confirming option D. Step 5: Option C is wrong. Σr³ = (Σr)² only equals Σr when Σr = 1, i.e., only for n=1. For n=2: Σr³=9, Σr=3, they are not equal. So saying they are always equal is false.