Math, asked by naira6262, 1 year ago

Find the sum of the series 2+4(3)²+6(5)²+8(7)².....+22(21)²

Answers

Answered by Selvasiddharth
15
Sum of the series =

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

Pls mark my answer as brainliest answer
Answered by amitnrw
1

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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