Question

2(1)²+4(3)²+6(5)²+......+22(21)² find the sum

Answer 1

Answer:

30932

Step-by-step explanation:

Use the well-known formulas:

• 1² + 2² + ... + n² = n (n+1) (2n+1) / 6
• 1³ + 2³ + ... + n³ = n² (n+1)² / 4

2(1)² + 4(3)² + 6(5)² + ... + 22(21)²

= (1+1)1² + (1+3)3² + (1+5)5² + ... + (1+21)21²

= 1² + 3² + 5² + ... + 21²

 + 1³ + 3³ + 5³ + ... + 21³

= ( 1² + 2² + 3² + ... + 21² ) - ( 2² + 4² + ... + 20² )

 + ( 1³ + 2³ + 3³ + ... + 21³ ) - ( 2³ + 4³ + ... + 20³ )

= (21)(22)(43) / 6  - 4×( 1² + 2² + ... + 10² )

 + (21²)(22)² / 4  - 8×( 1³ + 2³ + ... + 10³ )

= 3311 - 4 × (10)(11)(21) / 6

 + 53361 - 8 × (10²)(11²) / 4

= 3311 - 1540 + 53361 - 24200

= 1771 + 29161

= 30932