How do you find the sum of the series $3 i$ $3i$ from i=1 to i=6?

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Answer 1 · 2016-10-23T16:53:53

The sum of the series is 63.

Use the summation property $n \sum i c a_{i} = c n \sum i a_{i}$ $sum_i ^n ca_i = csum_i ^n a_i$ , where c is a constant.

$6 \sum 1 3 i = 3 \cdot 6 \sum 1 i$ $sum_1 ^6 3i=3*sum_1 ^6 i$

Use the summation property $n \sum x = 1 x = \frac{n (n + 1)}{2}$ $sum_(x=1) ^n x = frac{n(n+1)}[2}$

$3 \cdot 6 \sum 1 i = 3 \cdot \frac{6 (6 + 1)}{2} = 63$ $3* sum_1 ^6 i= 3 *frac{6(6+1)}{2} = 63$

Alternatively, you could substitute i =1, i=2, i=3,...i=6 into 3i and then add.

$3 (1) + 3 (2) + 3 (3) + 3 (4) + 3 (5) + 3 (6) = 63$ $3(1) +3(2)+3(3)+3(4)+3(5)+3(6) = 63$

How do you find the sum of the series 3i3i from i=1 to i=6?