Find the value of:
5Cfrac%7B1%7D%7B100%5E2%7D%29" id="TexFormula1" title="(1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{100^2})" alt="(1-\frac{1}{2^2})(1-\frac{1}{3^2})...(1-\frac{1}{100^2})" align="absmiddle" class="latex-formula"> help ASAP
2 answers:
<h3>Answer: 101/200</h3>
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Explanation:
Each term multiplied out is of the form 1 - 1/(k^2) which we can write as 1^2 - (1/k)^2
Note we have something in the form a^2 - b^2 with a = 1 and b = 1/k
So
a^2 - b^2 = (a-b)(a+b)
1^2 - (1/k)^2 = (1 - 1/k)(1 + 1/k)
1^2 - (1/k)^2 = [ (k - 1)/k ] * [ (k + 1)/k ]
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Let P = [ (k - 1)/k ] * [ (k + 1)/k ] represent the kth term we are dealing with
Then let Q be the term just after term P, so it's the (k+1) term
To form term Q, replace every k with k+1
Q = [ (k+1 - 1)/(k+1) ] * [ (k+1 + 1)/(k+1) ]
Q = [ k/(k+1) ] * [ (k+2)/(k+1) ]
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Let's multiply P and Q to find out what happens . See figure 1 in the attached images below.
As figure 1 shows, we see that the middle terms cancel out. This will happen when we multiply P*Q*R, where R is the term just after Q. Extend this idea out as far as you want and each time the middle terms will pair up and cancel.
All that's left are the first and last terms, which are
1 - 1/2 = 1/2
and
1 + 1/100 = 101/100
They multiply to 101/200
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Figure 2 (also attached as an image) shows a longer way to do the problem. Basically I expanded out a bunch of terms to multiply, and then show how we have various terms pair up and cancel, in basically the same way as explained above. What's left are the first and last terms 1/2 and 101/100 which multiply to 101/200.
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