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so I have this series negative five-thirds plus 25 over 6 minus 1 25 over 9 plus and it just keeps going on and on and on forever so this right over here is an infinite sum or an infinite series and what I want you to do right now is to pause this video and try to express this infinite series using Sigma notation so I'm assuming you've given a go at it so let's just look at each term of this series and let's see if we can express it with kind of an ever increasing index so the first thing that might jump out at you is this oscillating this oscillating sign that's happening right over here and whenever we see an oscillating sign that's a pretty good idea that we could kind of view this as negative 1 to the N where n is our index so for example that right over there is negative 1 to the first power that is this right over here this right over here is negative 1 to the second power that right over there is negative 1 to the third power so it looks like the sign is being is being defined by raising negative 1 to the index now let's look at the other parts of of these terms right over here so we have 5 then we have 25 then we have 125 so these are the powers of 5 so this right over here is 5 to the first power this right over here is 5 to the second power this right over here is 5 to the third power so this part we're raising 5 to our index notice 1 1 2 2 3 3 and then finally let's look at this we have 3 6 & 9 so this literally if our index here is 1 this is 3 times 1 if our index here is 2 this is 3 times 2 4 index here is 3 this is 3 times 3 so this is 3 times 1 that is 3 times 2 let me write it this way 3 times 2 that right over there is 3 times 3 so this sets us up pretty pretty well to write this in Sigma notation so let's write it over here just so we can we can compare so let me so let me give myself some real estate to work with so we could write this as the sum as I'll do it in yellow as the sum so this is our Sigma as the sum we can start our index we can start our index n at 1 from N equals 1 and we're going to keep going on and on forever we just keep going on and on forever and so it's negative 1 to the nth power negative 1 to the N power times 5 to the nth 5 to the N over notice 5 to the N over 3 n over 3 n is going to be equal to this and you can verify when N equals 1 its negative 1 to the nth power I'm sorry negative 1 to the first power which is negative 1 times 5 to the first power just 5 over 3 times 1 and we can do that for each successive term and so we're all we're all done