2+-3+4+-5+6 in sigma notation?
well, let's start simple. how would you express that if every number were positive? i.e. 2+3+4+5+6?
oh sorry my computer froze. If all the numbers were positive? +1
right, how would you write 2+3+4+5+6 in sigma notation?
um is that possible to do on a computer?
well you can just describe the lower and upper limits of the sigma and then the expression inside
OK well I'm new to this but lower would be 1 and upper 5 right?
are you learning from a book or some other reference that describes how to use sigma notation?
my teacher? who isn't here right now. Oh and the expression would be n+1
\[\sum_{1}^{5} n+1\]
yes, that's one way of doing it. ok, now we need to add something to that expression so that every other one is negative, any ideas on how to do that?
well in order to have the next number be negative, you have to be adding a negative number right?
right. so how can we turn every other number negative? i'll give you a hint ... you want to multiply the first number by 1, the second by -1, the third by 1 again, etc.
a hint? so you do know the answer...OK.
well what you just said is the part that confuses me. How can you multiple some numbers by -1, and others by +1? It's all 1 function right?
yes, i just thought you'd want to learn instead of only getting the answer
haha yeah I guess
It's just the last problem so I'm anxious
multiplying by -1 will alternate between positives and negatives
it is all one function but, i'll give you another hint ... what's (-1) squared?
1...
and what's -1 cubed?
-1
..oh.
and to the 4th power?
there you go, you cracked it. now put that together with your other expression.
yeah. got it.
excellent
thank you!
you're welcome, thanks for learning instead of just asking for the answer.
\[\sum_{1}^{5}\][(-1)^(n+1)]*(n+1)
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