Question

Write 3 − 5 + 7 − 9 + 11 − 13 + 15 in sigma notation.

Answer 1

+3
-(3+2)
+(3 + 2(2))
-(3 +3(2))
+(3 + 4(2))
...

Ideas?

Answer 2

6+6?

Answer 3

you have consecutive odd numbers
general odd number looks like \(2x-1\) for \(k=1,2,3,...\)
in this case you are starting at \(3\) and not 1, so maybe you can use \(2k+1\) instead
now you have to make the signs alternate \(-,+,-,+,-...\)
trick to doing that is two write (\(-1)^k\) which will be \(+1\) if \(k\) is even and \(-1\) if \(k\) is odd

Answer 4

in sigma notation it will look something like
\[\sum_{k=1}^6(-1)^{k+1}(2k+1)\]