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Let  S = 1^2/(1.3) + 2^2/(3.5) + 3^2/(5.7) + 4^2/(7.9) +............+ 500^2/(999.1001).

What is the sum of S?  Thank you for help.

 Jul 24, 2018
 #1
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Let S = 1^2/(1.3) + 2^2/(3.5) + 3^2/(5.7) + 4^2/(7.9) +.....+ 500^2/(999.1001)
Each term of S is of the form:n^2 /[(2n - 1)(2n + 1)]=1/4[n/(2n-1)+ n/(2n+1)]. Thus:
S = 1/4*∑[n/(2n-1)+ n/(2n+1)], n= 1 to 500
4S =∑[n/(2n-1)+ n/(2n+1)], n= 1 to 500=1/1+1/3+2/3+2/5+3/5+3/7+......+500/999+500/999. And since:
[n/(2n-1)+ (n+1)/(2n+1)] = 1, we can combine all such terms to get: 4S = 1 + 499 + 500/1001 =501,000/1,001, and:
S =125,250/1,011 =125 +125/1,011

 Jul 25, 2018
edited by Guest  Jul 25, 2018
 #2
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Let  S = 1^2/(1.3) + 2^2/(3.5) + 3^2/(5.7) + 4^2/(7.9) +............+ 500^2/(999.1001).

What is the sum of S?

 

\(\begin{array}{|rcll|} \hline S &=& \dfrac{1^2}{1\cdot 3} + \dfrac{2^2}{3\cdot 5} + \dfrac{3^2}{5\cdot 7} + \dfrac{4^2}{7\cdot 9} + \ldots + \dfrac{500^2}{999\cdot 1001} \\\\ S &=& \sum \limits_{n=1}^{500} \dfrac{n^2}{(2n-1)(2n+1)} \\\\ && \boxed{\dfrac{1}{(2n-1)(2n+1)}=\dfrac12\cdot \left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right)} \\\\ S &=& \sum \limits_{n=1}^{500} n^2\dfrac12 \left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right) \\\\ S &=& \dfrac12 \sum \limits_{n=1}^{500} n^2 \left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right) \\\\ S &=& \dfrac12 \sum \limits_{n=1}^{500} \left( \dfrac{n^2}{2n-1} - \dfrac{n^2}{2n+1} \right) \\\\ 2S &=& \sum \limits_{n=1}^{500} \left( \dfrac{n^2}{2n-1}\right) - \sum \limits_{n=1}^{500} \left( \dfrac{n^2}{2n+1} \right) \\\\ 2S &=& \dfrac{1^2}{1\cdot2-1} + \sum \limits_{n=2}^{500} \left(\dfrac{n^2}{2n-1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{2\cdot 500 + 1} \\\\ 2S &=& 1 + \sum \limits_{n=2}^{500} \left(\dfrac{n^2}{2n-1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{(n+1)^2}{2(n+1)-1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{(n+1)^2}{2n+1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{(n+1)^2-n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{n^2+2n+1-n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{2n+1}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(1 \right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + 499 -\dfrac{500^2}{1001} \\\\ 2S &=& 500-\dfrac{500^2}{1001} \\\\ 2S &=& 500 \cdot \left(1-\dfrac{500}{1001} \right) \\\\ S &=& 250 \cdot \left(\dfrac{1001-500}{1001} \right) \\\\ S &=& \dfrac{250\cdot 501}{1001} \\\\ \mathbf{S} & \mathbf{=} & \mathbf{\dfrac{125250}{1001}} \\ \hline \end{array}\)

 

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 Jul 25, 2018

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