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OpenStudy (anonymous):
Was wondering, I think this statement (in the comments) is true.
I think it is, and I am pretty sure.
Confrim please....
11 years ago
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OpenStudy (anonymous):
\[\LARGE c_{_1} \times \sum_{n=c_{_2}}^{c_{_3}}f(n)= \sum_{n=c_{_2}}^{c_{_3}} c_{_1} \times f(n)\]
(where \[\large c_{_1}~,~ c_{_2}~,~ c_{_3},\] are just 3 numbers that can be, but not necessarily equal to each other)
and assuming that
\[ \large c_{_1}\ne 0,~~1\]
11 years ago
OpenStudy (anonymous):
\[\large {\rm and} {~~c_{_3}} {\rm~~is~~a~~nat ural~~number. }\]
11 years ago
OpenStudy (anonymous):
whoa
11 years ago
OpenStudy (anonymous):
what?
11 years ago
OpenStudy (anonymous):
the notations?
11 years ago
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OpenStudy (anonymous):
YA
11 years ago
OpenStudy (anonymous):
plus ur rt
11 years ago
OpenStudy (anonymous):
basic summation FORMULA
11 years ago
OpenStudy (anonymous):
I like to be precise....
11 years ago
OpenStudy (anonymous):
yes, that is true, then.
11 years ago
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OpenStudy (anonymous):
hmm.tru dat
11 years ago
OpenStudy (anonymous):
but you are saying, it's something I should have known
11 years ago
OpenStudy (anonymous):
there's also a product formula...I think I forgot
11 years ago
OpenStudy (anonymous):
we all shud know these.....
11 years ago
OpenStudy (anonymous):
surely:)
I was just trying to appeal to my own logic
11 years ago
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OpenStudy (anonymous):
hmm lol man
11 years ago
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