Original subtitles

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Today are going to be talking about how to calculate the first several terms of a sequence.

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And in this particular problem we've been given a 7th term of the sequence which is a seben is equal

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to 1 plus the quantity negative one half to the end.

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Now when we're asked to calculate the first terms of the sequence and the A7 term is this simple.

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All we need to do is plug in values of n.

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Starting with N equals 1 and simplifying to find the value of each term in the sequence.

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So for example we'll say here and is equal to 1.

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We're going to get the a sub one term of the sequence and that's going to be plugging in one for and.

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Will get one.

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Plus the quantity negative one half to the first power and then we're going to get a sub one is equal

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to positive one half or point 5 0 0 0.

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Often when we're capturing the first terms of the sequence we're asked to calculate to four decimal

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places or something like that.

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So we can go ahead and express it that way as well.

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And then we just follow the same process for the next terms in the sequence we say any equals two is

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going to give us a sub 2 which will be 1 plus negative 1 half squared.

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That's going to give us a sub 2 is equal to one plus one fourth or five fourths which is the same as

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1 point 2 5 0 0.

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We can continue on like this where we get an equals three.

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That's the ace of three term.

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That's one plus quantity negative one half to the third which gives us a sub 3 is equal to 1 minus 1

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eighth or just seven eighths which is the same as point 8 7 5 0.

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And I will keep going with all of the tedious writing but what you'll find is that if you keep going

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let's say to the first six terms of the sequence you'll get a sub 4 equals 17 over 16 or one point zero

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six to five.

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You'll get a sub 5 is equal to 31 over 32 which is also equal to point 9 6 8 8.

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And then you'll get a sub 6 is equal to 65 over 64 which is also one point 0 1 5 6 and you can just

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keep going like that if you want to try to use these terms to draw a conclusion about the limit of the

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sequence as and approaches infinity.

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You can go ahead and try to plot these values on a coordinate system and see if you can make a guess

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about the value they're approaching.

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I've gone ahead and done that already anyway.

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And what you can see is that if we plot these values and this is values for and here we found an equals

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1 2 3 and then 4 5 6.

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So this is the ace of one value here we found that with point five.

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So we plotted that point here.

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At one point five and we just kept going.

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And what you see is that these values oscillate back and forth across the line y equals one but they

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get closer and closer to that line.

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So the conclusion that we can then try to draw is that the limit as and goes to infinity of the sequence

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A saw then this entire sequence here is equal to 1 because as and goes farther and farther out to the

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right these points will approach the lying White was one more get closer and closer and closer to it

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without ever reaching it.

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So the limit is and approaches infinity of the entire sequence is one.