value of `1+(1^(3)+2^(3))/(1+2)+(1^(3)+2^(3)+3^(3))/(1+2+3)+……+(1^(3)+2^(3)+….15^(3))/(1+2+…+15)-(1)/(2)(1+2+….+15)` is

Options

A. 620
B. 660
C. 1240
D. 1860

Answer

A

Explanation

Given scries,
`S = 1 + (1^(3) + 2^(3))/(1 + 2) + (1^(3) + 2^(3) + 3^(3))/(1 + 2 + 3) + ..+ (1^(3) + 2^(3) + 3^(3) + ...+ 15^(3))/(1 + 2 + 3 + ...+ 15) - (1)/(2) (1 + 2 + 3 + ..+ 15)`
`= S_(1) - S_(2)` (let)
where,
`S_(1) = 1 + (1^(3) + 2^(3))/(1 +2) + (1^(3) + 2^(3) + 3^(3))/(1 + 2 + 3) + ..+ (1^(3) + 2^(3) + 3^(3) + ...+ 15^(3))/(1 + 2 + 3 + ...+ 15)`
`= underset(n=1)overset(15)sum (1^(3) + 2^(3) + ...+ n^(3))/(1 + 2 + ...+ n) = underset(n =1)overset(15)sum (((n(n+1))/(2))^(2))/((n(n+1))/(2))`
`[ :' underset(r =1)overset(n)sum r^(3) = ((n(n+1))/(2))^(2) and underset( r =1)overset(n)sum r = (n (n+1))/(2)]`
`= underset(n=1)overset(15)sum (n(n+1))/(2) = (1)/(2) underset(n=1)overset(n)sum (n^(2) + n)`
`= (1)/(2) [(15 xx 16 xx 31)/(6) + (15 xx 16)/(2)]`
`[ :' underset(r=1)overset(n)sum r^(2) = (n(n+1) (2n +1))/(6)`
`= (1)/(2) [(5 xx 8 xx 31) + (15 xx 8)]`
`= (5 xx 4 xx 31) + (15 xx 4)`
`= 620 + 60 = 680`
and `S_(2) = (1)/(2) (1 + 2 + 3 + ...+ 15)`
`= (1)/(2) xx (15 xx 16)/(2) = 60`
Therefore, `S = S_(1) - S_(2) = 680 - 60 = 620`