The sum up to 10 terms of the series `(3 xx 1^(3))/(1^(2)) + (5 xx (1^(3) + 2^(3)))/(1^(2) + 2^(2)) + (7 xx (1^(3) + 2^(3) + 3^(3)))/(1^(2) + 2^(2) + 3^(2)) +`....

Options

A. 680
B. 600
C. 660
D. 620

Answer

C

Explanation

Given series is
`(3 xx 1^(3))/(1^(2)) + (5xx (1^(3) + 2^(3)))/(1^(2) + 2^(2)) + (7 xx (1^(3) + 2^(3) + 3^(3)))/(1^(2) + 2^(2) + 3^(3)) +`...
So, nth term
`T_(n) = ((3 + (n -1) 2) (1^(3) + 2^(3) + 3^(3) ...+ n^(3)))/(1^(2) + 2^(2) + 3^(2) + ...+ n^(2))`
`= ((2n + 1) xx ((n (n +1))/(2))^(2))/((n (n +1) (2n +1))/(6))`
`[ :' underset(r=1)overset(n)Sigma r^(3) = [(n(n+1))/(2)]^(2) and underset(r =1)overset(n)Sigma r^(2) = (n(n+1) (2n+1))/(6)]`
So, `T_(n) = (3n (n +1))/(2) = (3)/(2) (n^(2) + n)`
Now, sum of the given series upto n terms
`S_(n) = Sigma T_(n) = (3)/(2) [Sigma n^(2) + Sigma n]`
`= (3)/(2) [(n (n+1) (2n+1))/(6) + (n(n+1))/(2)]`
`:. S_(10) = (3)/(2) [(10 xx 11 xx 21)/(6) + (10 xx 11)/(2)]`
`= (3)/(2) [(5 xx 11 xx 7) + (5xx 11)]`
`= (3)/(2) xx 55 (7 + 1) = (3)/(2) xx 55 xx 8 = 3 xx 55 xx 4`
`= 12 xx 55 = 660`