Find the sum of the 2i - 9 from i = 3 to 10.
Solution:
\(\sum_{i=3}^{10}2i-9=\sum_{i=1}^{10}2i-9-\sum_{i=1}^{2}2i-9\)
First let us consider \(\sum_{i=1}^{10}2i-9\)
To fit the summation rules let us split it into smaller summations
\(\sum_{i=1}^{10}2i-9=2\sum_{i=1}^{10}i+\sum_{i=1}^{10}-9\)
For summation of a polynomial with degree 1
\(\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\)
By substituting the values, we get
= \((2)(\frac{10(10+1)}{2})\)
On further simplification
= 2 (10 × 11)/2
= 110
The summation of a constant formula is \(\sum_{i=1}^{n}c=cn\)
By substituting the values, we get
= (-9)(10)
= -90
So we get,
\(\sum_{i=1}^{10}2i-9=2\sum_{i=1}^{10}i+\sum_{i=1}^{10}-9\)
= 110 - 90
= 20
To fit the summation rules let us split it into smaller summations
\(\sum_{i=1}^{2}2i-9=2\sum_{i=1}^{2}i+\sum_{i=1}^{2}-9\)
For summation of a polynomial with degree 1
\(\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\)
By substituting the values, we get
= \((2)(\frac{2(2+1)}{2})\)
On further simplification
= 2(2 × 3)/2
= 6
The summation of a constant formula is \(\sum_{i=1}^{n}c=cn\)
By substituting the values
= (-9)(2)
= -18
So we get,
\(\sum_{i=1}^{2}2i-9=2\sum_{i=1}^{2}i+\sum_{i=1}^{2}-9\)
= 6 - 18
= -12
Here,
\(\sum_{i=3}^{10}2i-9=\sum_{i=1}^{10}2i-9-\sum_{i=1}^{2}2i-9\)
= 20 + 12
= 32
Therefore, the sum of \(\sum_{i=3}^{10}2i-9\) is 32.
Find the sum of the 2i - 9 from i = 3 to 10.
Summary:
The sum of \(\sum_{i=3}^{10}2i-9\) is 32.
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