Question 951201
The multiples of 5 can be represented by the expression {{{5n}}} where {{{n}}} is a whole number


Ex: if {{{n = 2}}}, then {{{5n = 5(2) = 10}}}


How do we alternate the signs? We introduce the term {{{(-1)^n}}}. If n is even, then {{{(-1)^n=1}}} (ie positive 1). If n is odd, then {{{(-1)^n=-1}}}. There are no other outcomes for this piece. So this is why {{{(-1)^n}}} is the perfect thing to tack on to help alternate the signs.


The nth term is {{{(-1)^n*(5n)}}}. Notice how I am multiplying {{{(-1)^n}}} by {{{5n}}}


To confirm this, let's plug in n = 2 to get {{{(-1)^n*(5n) = (-1)^2*(5*2) = 1*10 = 10}}}. 


Now let's plug in n = 3 to get {{{(-1)^n*(5n) = (-1)^3*(5*3) = -1*15 = -15}}}. 


I'll let you check the others.


The next term is -25 as I'm sure you've already figured that out. Let's use n = 5 to confirm


{{{(-1)^n*(5n) = (-1)^5*(5*5) = -1*25 = -25}}}. 

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