Question

Find sun of first 15 term of series `1+6+(9(1^2+2^2+3^2))/(7)+(12(1^2+2^2+3^2+4^2))/(9)+`?. (a) `7620` (b) `7280` (c) `7820` (d) `7067`
A. 7510
B. 7820
C. 7830
D. 7520

Answer 1

Correct Answer - B
General term of the given series is
`T_(r) = (3r (1^(2) + 2^(2) + ...+ r^(2)))/(2r +1) = (3r [r ( r +1) (2r +1)])/(6(2r +1))`
`= (1)/(2) (r^(3) + r^(2))`
Now, required sum `= underset(r =1)overset(15)sum T_(r) = (1)/(2) underset(r=1)overset(15)sum (r^(3) + r^(2))`
`= (1)/(2) {[(n (n +1))/(2)]^(2) + (n(n+1) (2n+1))/(6)}_(n=15)`
`= (1)/(2) {(n(n+1))/(2)[(n^(2) +n)/(2) + (2n +1)/(3)]}_(n =15)`
`= (1)/(2) {(n(n+1))/(2) ((3n^(2) + 7n +2))/(6)}_(n=15)`
`= (1)/(2) xx (15 xx 16)/(2) xx ((3 xx 225 + 105 + 2))/(6) = 7820`