Find the sum of the 2i-12 from i =7 to i = 16.
Solution:
\(\sum_{i=7}^{16}2i-12=\sum_{i=1}^{16}2i-12-\sum_{i=1}^{6}2i-12\)
First let us consider \(\sum_{i=1}^{16}2i-12\)
To fit the summation rules let us split it into smaller summations
\(\sum_{i=1}^{16}2i-12=2\sum_{i=1}^{16}i+\sum_{i=1}^{16}-12\)
For summation of a polynomial with degree 1
\(\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\)
By substituting the values
\((2)(\frac{16(16+1)}{2})\)
On further simplification
= 2(16 × 17)/2
= 272
The summation of a constant formula is
\(\sum_{i=1}^{n}c=cn\)
By substituting the values,
= (-12)(16)
= -192
So we get,
\(\sum_{i=1}^{16}2i-12=2\sum_{i=1}^{16}i+\sum_{i=1}^{16}-12\)
= 272 - 192
= 80
To fit the summation rules let us split it into smaller summations
\(\sum_{i=1}^{6}2i-12=2\sum_{i=1}^{6}i+\sum_{i=1}^{6}-12\)
For summation of a polynomial with degree 1
\(\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\)
By substituting the values
\((2)(\frac{6(6+1)}{2})\)
On further simplification
= 2(6 × 7)/2
= 42
The summation of a constant formula is
\(\sum_{i=1}^{n}c=cn\)
By substituting the values
= (-12)(6)
= -72
So we get,
\(\sum_{i=1}^{6}2i-12=2\sum_{i=1}^{6}i+\sum_{i=1}^{6}-12\)
= 42 - 72
= -30
Here,
\(\sum_{i=7}^{16}2i-12=\sum_{i=1}^{16}2i-12-\sum_{i=1}^{6}2i-12\)
= 80 + 30
= 110
Therefore, the sum is 110.
Find the sum of the 2i-12 from i=7 to i = 16.
Summary:
The sum of \(\sum_{i=7}^{16}2i-12\) is 110.
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