Difference between revisions of "2006 iTest Problems/Problem 31"
Rockmanex3 (talk | contribs) (Solution to Problem 31 -- Infinite Summation of a Rational Function) |
Rockmanex3 (talk | contribs) m (→Solution) |
||
Line 10: | Line 10: | ||
<cmath>\sum_{n=2}^\infty\dfrac{1}{n^2 - 1} + \dfrac{n}{n^4 + 2n^2 + 1 - n^2}</cmath> | <cmath>\sum_{n=2}^\infty\dfrac{1}{n^2 - 1} + \dfrac{n}{n^4 + 2n^2 + 1 - n^2}</cmath> | ||
<cmath>\sum_{n=2}^\infty\dfrac{1}{(n+1)(n-1)} + \dfrac{n}{(n^2 + n + 1)(n^2 - n + 1)}</cmath> | <cmath>\sum_{n=2}^\infty\dfrac{1}{(n+1)(n-1)} + \dfrac{n}{(n^2 + n + 1)(n^2 - n + 1)}</cmath> | ||
− | In the fraction <math>\frac{1}{(n+1)(n-1)}</math>, we can split the fraction as <math>\frac{1}{2(n-1)} - \frac{1}{2(n+1)}</math>, making <math>\sum_{n=2}^\infty\dfrac{1}{(n+1)(n-1)}</math> a [[telescoping]] series. Plugging in values results in <math>\frac12 - \frac16 + \frac14 - \frac18 + \frac16 - \ | + | In the fraction <math>\frac{1}{(n+1)(n-1)}</math>, we can split the fraction as <math>\frac{1}{2(n-1)} - \frac{1}{2(n+1)}</math>, making <math>\sum_{n=2}^\infty\dfrac{1}{(n+1)(n-1)}</math> a [[telescoping]] series. Plugging in values results in <math>\frac12 - \frac16 + \frac14 - \frac18 + \frac16 - \frac1{10} + \cdots = \frac12 + \frac14 = \frac34</math>. |
<br> | <br> |
Latest revision as of 20:47, 11 December 2018
Problem
The value of the infinite series can be expressed as where and are relatively prime positive numbers. Compute .
Solution
Notice that , and notice that the numerator contains . Also, notice that . We can rewrite the expression by factoring and splitting the fractions. In the fraction , we can split the fraction as , making a telescoping series. Plugging in values results in .
In the fraction , plugging in the first few values results in . The structure suggests that we could try to model the same fraction with a telescoping series.
Substituting for in results in . Also, because , we find that the expression is equal to , confirming that is a telescoping series. The expression equals .
Therefore, , so .
See Also
2006 iTest (Problems) | ||
Preceded by: Problem 30 |
Followed by: Problem 32 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10 |