The value of -r-1630r-2r-3 is equal to-

∑_r=16^30(r+2)(r 3) =∑_r=16^30(r^2 r 6) =∑_r=1^30(r^2 r 6) ∑_r=1^15(r^2 r 6) = (30 × 31 × 61/6 30 × 31/2 6 × 30) (15 × 16 × 31/6 15 × 16/2 6 × 15) =8810 1030 ...

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The value of $30 \sum r = 16 (r + 2) (r - 3)$ is equal to:

7775

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7785

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7770

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7780

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Solution

The correct option is D $7780$
$30 \sum r = 16 (r + 2) (r - 3)$
$= 30 \sum r = 16 (r^{2} - r - 6)$
$= 30 \sum r = 1 (r^{2} - r - 6) - 15 \sum r = 1 (r^{2} - r - 6)$
$= (\frac{30 \times 31 \times 61}{6} - \frac{30 \times 31}{2} - 6 \times 30) - (\frac{15 \times 16 \times 31}{6} - \frac{15 \times 16}{2} - 6 \times 15)$
$= 8810 - 1030$
$= 7780$

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The value of 30∑r=16(r+2)(r−3) is equal to:

The correct option is D 7780 30∑r=16(r+2)(r−3) =30∑r=16(r2−r−6) =30∑r=1(r2−r−6)−15∑r=1(r2−r−6) =(30×31×616−30×312−6×30)−(15×16×316−15×162−6×15) =8810−1030 =7780

The value of $30 \sum r = 16 (r + 2) (r - 3)$ is equal to: