The value of sumlimits_{r = 16}^{30} {(r + 2) | vedclass

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(D) We need to evaluate the sum $S = \sum\limits_{r = 16}^{30} {(r^2 - r - 6)}$.Using the summation properties,$S = \sum\limits_{r = 16}^{30} r^2 - \s

Vedclass · Accepted Answer

$7780$

Vedclass · Answer

(D) We need to evaluate the sum $S = \sum\limits_{r = 16}^{30} {(r^2 - r - 6)}$.Using the summation properties,$S = \sum\limits_{r = 16}^{30} r^2 - \sum\limits_{r = 16}^{30} r - \sum\limits_{r = 16}^{30} 6$.Recall that $\sum\limits_{r = 1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum\limits_{r = 1}^{n} r = \frac{n(n+1)}{2}$.$\sum\limits_{r = 16}^{30} r^2 = \sum\limits_{r = 1}^{30} r^2 - \sum\limits_{r = 1}^{15} r^2 = \frac{30(31)(61)}{6} - \frac{15(16)(31)}{6} = 9455 - 1240 = 8215$.$\sum\limits_{r = 16}^{30} r = \sum\limits_{r = 1}^{30} r - \sum\limits_{r = 1}^{15} r = \frac{30(31)}{2} - \frac{15(16)}{2} = 465 - 120 = 345$.$\sum\limits_{r = 16}^{30} 6 = 6 	imes (30 - 16 + 1) = 6 	imes 15 = 90$.Therefore,$S = 8215 - 345 - 90 = 7780$.

The value of $\sum\limits_{r = 16}^{30} {(r + 2)(r - 3)}$ is equal to

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