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Solution for finding the sum of an infinite sequence

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Introduction

Internal Assessment 1

Solution for finding the sum of a infinite sequence

The objective of this assignment is to find out the sum of infinite sequences image00.pngimage00.png, where

image44.png

image77.png

In this equation, image106.pngimage106.png is defined by the term number. For example, image120.pngimage120.png is the first term, whereas image00.pngimage00.png is the nth term. image01.pngimage01.png are various variables. In this equation, image20.pngimage20.png which is defined by image26.pngimage26.png. Likewise 3! is defined by 3image33.pngimage33.png, but there is an exception image39.pngimage39.png

First I will break down the equation so that it will be easier for me to find out he formula. I will examine the image07.pngimage07.png defined as the sum of the first image00.pngimage00.png. For example, image51.pngimage51.png, image61.pngimage61.png. I am first going to use this equation image69.pngimage69.png, where image22.pngimage22.png, image31.pngimage31.png, and where image19.pngimage19.png is image92.pngimage92.png. So it should look like this:

image101.png

I will find out the image07.pngimage07.png for image92.pngimage92.png.

image113.png

In order to find image07.pngimage07.png, I used my TI-84 Plus to figure this out. I will plug in the equation                  image21.pngimage21.png, where the image28.pngimage28.png is 1, image42.pngimage42.png is 2 and image19.pngimage19.png value is from 0 to 10. The method is shown in the appendix.

I came up with:

image114.png

image115.png

image116.png

image117.png

image118.png

image119.png

image121.png

image122.png

image123.png

image124.png

image125.png

In order to check if I got it right, I used Microsoft Excel 2010. The method is shown in the appendix.

image126.jpg

After seeing that my result matches the results in Excel, I decided to then find out the sum.

...read more.

Middle

In order to find image07.pngimage07.png, I will only use Microsoft Excel 2010 since I know that it gave me the right result before. I will plug in the equation image21.pngimage21.png, where the image28.pngimage28.png is 1, image29.pngimage29.pngis 3 and image19.pngimage19.png value is from 0 to 10. The method is shown in the appendix.

image30.jpg

To find the results, I just subbed in 3 instead if image31.pngimage31.png. Now I will graph it using the same process:


image32.png

By looking at this graph, I can say that when image19.pngimage19.png value is past 10, image07.pngimage07.png is 3 and it remains constant. This suggests that when image19.pngimage19.png approaches image34.pngimage34.png (infinity), sum of the infinity image35.pngimage35.png value is 3.

After looking at the two sum, whenimage36.pngimage36.png, and when image23.pngimage23.png, I think that the general term is

image37.png

Where image38.pngimage38.png is the sum of the infinite numbers.  I came up with this formula because we know,

  • For image31.pngimage31.png, when image19.pngimage19.png kept increasing, the sum of the image19.pngimage19.png value remained the same and that’s why image40.pngimage40.png. We can say that image41.pngimage41.png instead of image40.pngimage40.png because image19.pngimage19.png can go up to image34.pngimage34.png and on the other hand image42.pngimage42.png is equivalent to 2.
  • For image23.pngimage23.png, when image19.pngimage19.png kept increasing, the sum of the image19.pngimage19.png value remained the same and that’s why image43.pngimage43.png. We can say that image41.pngimage41.png instead of image43.pngimage43.png because image19.pngimage19.png can go up to image34.pngimage34.png and on the other hand image42.pngimage42.png is equivalent to 3.

Thus, image37.pngimage37.png

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Conclusion

image95.jpg

After I pressed ENTER key, I got this result –

image96.jpg

This suggest there is no result when image93.pngimage93.png. The reason behind this is that, when we punch in image97.pngimage97.png or image98.pngimage98.pngimage99.png, we get the same error. As we cannot solve it.

Now I am going to let aimage100.pngimage100.png, image72.pngimage72.png and image94.pngimage94.png in the equation image71.pngimage71.png.

I will use my TI-84 Plus in order to find out the situation. It looks like this –

image102.jpg

So this means, when I use image42.pngimage42.png as a negative number it will show some results. I will check if the result shown above different from the equation image103.pngimage103.png.

image104.png

This means, that when image31.pngimage31.png it is equal to when image105.pngimage105.png. But what happens when image42.pngimage42.png is a negative number and image28.pngimage28.png is a odd number. For example, when image107.pngimage107.png.

I will let image105.pngimage105.png, image57.pngimage57.png and image94.pngimage94.png in  the equation image71.pngimage71.png.

 I will again use my TI-84 Plus to find out the result –

image108.jpg

This is what I get when I hit enter –

image109.jpg

This simple means that you cannot solve a equation when there is a negative number. For example,

image110.png

In conclusion, when image42.pngimage42.png is a negative number, image28.pngimage28.png had to be an even number to change the negative sign to positive sign.

In short, image111.pngimage111.png, image112.pngimage112.png, image28.pngimage28.png can be both positive and negative number, image42.pngimage42.png can also be both negative and positive number but only when image28.pngimage28.png is an even number.

...read more.

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