Find the sum of the series 2+4(3)²+6(5)²+8(7)².....+22(21)²
Answers
2 + 4(3)^2 + 6(5)^2 + 8(7)^2 + 10(9)^2 + 12(11)^2 + 14(13)^2 + 16(15)^2 + 18(17)^2 + 20(19)^2 + 22(21)^2
=2+36+150+392+810+1452+2366+3600+5202+7220+9702
=30932
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Given : 2+4(3)²+6(5)²+8(7)².....+22(21)²
To find : Sum of Series 2+4(3)²+6(5)²+8(7)².....+22(21)²
Solution:
2+4(3)²+6(5)²+8(7)².....+22(21)²
= 2(1)² + 4(3)²+6(5)²+8(7)².....+22(21)²
Tₙ = 2n(2n-1)²
=> Tₙ = 2n(4n² - 4n + 1)
=> Tₙ = 8n³ - 8n² + 2n
=> Tₙ = 2(4n³ - 4n² + n)
∑Tₙ = 2 * (4 ∑n³ - 4∑n² + ∑n)
=> ∑Tₙ = 2 * (4 ((n)(n + 1)/2)² - 4n(n+1)(2n+1)/6 + n(n+1)/2)
=> ∑Tₙ = 2 n(n+1) ( (n)(n + 1) - 2 (2n+1)/3 + 1/2)
=> ∑Tₙ = 2 n(n+1) ( 6(n)(n + 1) - 4 (2n+1) + 3)/6
=> ∑Tₙ = n(n+1) (6n² + 6n - 8n - 4 + 3)/3
=> ∑Tₙ = n(n+1) (6n²- 2n -1 )/3
2+4(3)²+6(5)²+8(7)².....+22(21)²
n = 1 to 11
∑Tₙ
= 11 (11 + 1)( 6 * 11² - 2*11 - 1)/3
= 11 * 12 ( 726 - 22 - 1)/3
= 44 (703)
= 30932
30932 is Sum of Series 2+4(3)²+6(5)²+8(7)².....+22(21)²
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