How do you find the partial sum of $100 \sum r = 0 \frac{8 - 3 r}{16}$ $sum_(r=0)^(100)(8-3r)/16$ ?

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Answer 1 · 2017-12-08T12:27:26

$100 \sum r = 0 = - 896.375$ $sum_(r=0)^100= -896.375$

$100 \sum r = 0 = \frac{8 - 3 r}{16} = 100 \sum r = 0 \frac{8}{16} - \frac{3 r}{16}$ $sum_(r=0)^100=(8-3r)/16= sum_(r=0)^100 8/16 - (3r)/16$

$= 100 \sum r = 0 \frac{1}{2} - \frac{3}{16} \cdot 100 \sum r = 0 r$ $=sum_(r=0)^100 1/2 -3/16* sum_(r=0)^100 r$ or

$[n \sum r = 0 = \frac{n (n + 1)}{2}]$ $[sum_(r=0)^n=(n(n+1))/2]$

$100 \sum r = 0 = \frac{1}{2} \cdot 101 - \frac{3}{16} \cdot \frac{100 \cdot 101}{2}$ $sum_(r=0)^100=1/2*101 - 3/16* (100*101)/2$

$100 \sum r = 0 = 50.5 - \frac{3}{16} \cdot 5050$ $sum_(r=0)^100=50.5 - 3/16* 5050$ or

$100 \sum r = 0 = 50.5 - 946.875 = - 896.375$ $sum_(r=0)^100=50.5 - 946.875= -896.375$ [Ans]

How do you find the partial sum of sum_(r=0)^(100)(8-3r)/16100∑r=08−3r16sum_(r=0)^(100)(8-3r)/16 ?