How do you express the sum #5-10+20-40+80-.....# in sigma notation?

#sum = 5-10+20-40+80-.....#

Here we have a sequence of terms with alternating sign beginning positive.

To achieve this we need a term #(-1)^(1+1)# for #i# from #1 to n#

Only considering the absolute values of the terms, we have sequence where #a_n = 2*a_"n-1"# and #a_1 = 5#

To achieve this we need a term #5*2^(i-1)# for #i# from #1 to n#

Combining these results we have our sum in sigma notation:

#sum_"i=1"^n" (-1)^(i+1) *5*2^(i-1)#

One form of this is:

#5 cdot sum_(n=0)^(N) (-1)^(n) 2^n#

Perhaps a simpler form is:

#5 cdot sum_(n=0)^(N) (-2)^n#

Notice how the starting term is #5#, and it gets doubled with an alternating sign each time. A starting point is:

#5 cdot sum_(n=0)^(N) 2^n = 5[2^0 + 2^1 + 2^2 + 2^3 + . . . ]#

#= 5 + 10 + 20 + 40 + . . . #

Now, to add the alternating sign, we just need to put in a #(-1)^(n)#, as for #n = 0, 1, 2, . . . #,

#{(-1)^(n)} = (-1)^(0), (-1)^(1), (-1)^(2), . . . #

#= 1, -1, 1, . . . #

This gives:

#color(blue)(5 cdot sum_(n=0)^(N) (-1)^(n) 2^n)#

#= 5[(-1)^(0) 2^(0) + (-1)^(1) 2^(1) + (-1)^(2) 2^(2) + (-1)^(3) 2^(3) + (-1)^(4) 2^(4) + . . . ]#

#= 5[1 - 2 + 4 - 8 + 16 - . . . ]#

#= color(blue)(5 - 10 + 20 - 40 + 80 - . . . )#