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The value of $\sum_{r=16}^{30}(r+2)(r-2)$ is equal to
Option: 1
7785

Option: 2
8155

Option: 3
7775

Option: 4
8100

Answers (1)

best_answer

$\sum_{r=16}^{30}\left ( r+2 \right )\left ( r-2 \right )$

$=\sum_{r=16}^{30}\left ( r^2 - 4 \right )$

$=(16^{2}+17^{2}+18^{2} - - - -\30^{2})-4\times 15$

Now $1^{2}+2^{2}+3^{2}+- - - - -+30^{2}=\frac{30\times 31\times 61}{6}$

So that $16^{2}+17^{2}+ - - - - - - 30^{2}$

$=\frac{30\times 31\times 61}{6} -\left ( 1^{2} +2^{2}+..........+15^{2}\right )$

$=\frac{30\times 31\times 61}{6} -\frac{15\times 16\times 31}{6}$

$=\frac{31}{6}\left [ 1830-240 \right ]$

$=\frac{31}{6}\times 1590$

$=\frac{31}{6}\times 265$

$=8215$

So $\sum_{r=16}^{30}(16^{2}+17^{2}+ - - - - - - 30^{2})-60$

$=8215-60$

$=8155$

Posted by

Ritika Harsh

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