How to find the partial fractions for this expression?

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The discussion focuses on finding the partial fractions for the expression ((n+1)*(sqrt(n)) - n*(sqrt(n+1))) / (n*(n+1)). The attempted solution resulted in A = 0 and B = 0, indicating a misunderstanding of the problem. It was clarified that the expression does not require partial fractions but rather needs to be rewritten and simplified into separate terms. The final answer derived from the correct approach is 1/sqrt(n) - 1/(sqrt(n+1)). Simplification of the expression is essential for accurate results.
ybhathena
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Homework Statement



Find the partial fractions for this expression.

(((n+1)*(sqrt(n)) - n*(sqrt(n+1))) / (n*(n+1)))

The Attempt at a Solution



The final answer is 1/sqrt(n) - 1/(sqrt(n+1))

My work:

A/n - B/(n+1) = n*sqrt(n+1) - (n+1)*(sqrt(n))
I am subbing in n = -1 and n = 0 to solve for A and B which usually works but in this case it is giving me A = 0 and B = 0, which means I am getting 0 as my partial fraction. Thank you for your help.
 
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ybhathena said:

Homework Statement



Find the partial fractions for this expression.

(((n+1)*(sqrt(n)) - n*(sqrt(n+1))) / (n*(n+1)))

The Attempt at a Solution



The final answer is 1/sqrt(n) - 1/(sqrt(n+1))

My work:

A/n - B/(n+1) = n*sqrt(n+1) - (n+1)*(sqrt(n))
I am subbing in n = -1 and n = 0 to solve for A and B which usually works but in this case it is giving me A = 0 and B = 0, which means I am getting 0 as my partial fraction. Thank you for your help.

Your expression doesn't require partial fractions, it requires re-writing as <br /> \frac{(n+1)\sqrt{n} - n\sqrt{n+1}}{n(n+1)} = \frac{(n+1)\sqrt{n}}{n(n+1)} -\frac{n\sqrt{n+1}}{n(n+1)} and further simplifications in each term.
 
Thank you very much !
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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