The sum of the series `1 + 2 xx 3 + 3 xx 5 + 4 xx 7+`...upto 11th term is

Options

A. 915
B. 946
C. 916
D. 945

Answer

B

Explanation

Given series is
`1 + (2 xx 3) + (3 xx 5) + (4xx 7) +`...upto 11 terms.
Now, the rth term of the series is `a_(r) = r (2r -1)`
`:.` Sum of first 11-terms is
`S_(11) = underset(r=1)overset(11)sum r (2r -1) = underset(r=1)overset(11)summ (2r^(2) -r) = 2 underset(r=1)overset(11)sum r^(2) - underset(r =1)overset(11)sum r`
`= 2 (11 xx (11 +1) (2 xx 11 + 1))/(6) - (11 xx (11 + 1))/(2)`
`[ :' underset(r=1)overset(n)sum r^(2) = (n (n+1) (2n +1))/(6) and underset(r =1)overset(n)sum r = (n(n +1))/(2)]`
`= ((11 xx 12 xx 23)/(3)) - ((11 xx 12)/(2))`
`= (11 xx 4 xx 23) - (11 xx 6) = 11 (92 -6) = 11 xx 86 = 946`