All Questions
17
questions
17
votes
2
answers
862
views
proof of $\sum\nolimits_{i = 1}^{n } {\prod\nolimits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i - x_j }}} } = 1$ [duplicate]
i found a equation that holds for any natural number of n and any $x_i \ne x_j$ as follows:
$$\sum\limits_{i = 1}^{n } {\prod\limits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i - x_j }}} } ...
7
votes
1
answer
244
views
Prove that $\frac{1}{1+a_1+a_1a_2}+\frac{1}{1+a_2+a_2a_3}+\cdots+\frac{1}{1+a_{n-1}+a_{n-1}a_n}+\frac{1}{1+a_n+a_na_1}>1.$
If $n > 3$ and $a_1,a_2,\ldots,a_n$ are positive real numbers with $a_1a_2\cdots a_n = 1$, prove that $$\dfrac{1}{1+a_1+a_1a_2}+\dfrac{1}{1+a_2+a_2a_3}+\cdots+\dfrac{1}{1+a_{n-1}+a_{n-1}a_n}+\dfrac{...
6
votes
4
answers
828
views
Summing reciprocal logs of different bases
I recently took a math test that had the following problem:
$$
\frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!}
$$
The sum is equal to 1. I ...
5
votes
3
answers
201
views
Prove $\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$
Could you explain the operation in the third step?
$$\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$$
It comes from the sumation $$\sum_{n=1}^\...
3
votes
2
answers
154
views
Fraction Sum Series
This question was asked in (selection) IMO for 8th graders.
$1/2 + 1/6 + 1/12+ 1/20 + 1/30 + 1/42 +1/56 + 1/72 + 1/90 + 1/110 +1/132$
I have noticed that it can be written as $1/(1*2) + 1/(2*3) +1/(...
3
votes
1
answer
354
views
Prove $\frac{n-1}{n}+\frac{n-1}{n}\frac{n-3}{n-2}+ \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4} + ... = \frac{n-1}{3}$ without induction
I found this identity using Maple. Is there a (simple) way to prove it without using induction? Using induction, the proof is quite easy.
Prove for odd $n$ that
$$\sum_{k=1}^{(n+1)/2}\prod_{j=0}^{k-1}...
3
votes
1
answer
106
views
Inequality $\frac{x_1^2}{x_1^2+x_2x_3}+\frac{x_2^2}{x_2^2+x_3x_4}+\cdots+\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1}+\frac{x_n^2}{x_n^2+x_1x_2}\le n-1$
Show that for all $n\ge 2$
$$\frac{x_1^2}{x_1^2+x_2x_3}+\frac{x_2^2}{x_2^2+x_3x_4}+\cdots+\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1}+\frac{x_n^2}{x_n^2+x_1x_2}\le n-1$$
where $x_i$ are real positive ...
3
votes
0
answers
283
views
Pull constant out of a summation of fractions
General problem
$$
\sum_{i=1}^n \frac{a_i + x}{b_i + x} = 0
$$
Is it possible for solve for $x$?
Some context
I've hit a road block in my derivation... At this point, I need to pull the model ...
2
votes
1
answer
81
views
How to prove $\frac{\prod_{k=1}^{31}{}\left(1+\frac{29}{k}\right)}{ \prod_{k=1}^{29}\left(1+\frac{31}{k}\right)}=1$ [closed]
Here is a fraction that seems to stump me as to how to work it out. Using a calculator I find out that the answer is 1, and the answer sheet proves me correct. However, if I want to solve future ...
1
vote
2
answers
98
views
$f(n)=\sqrt{2n+1+2\sqrt{n^2+n}}$, $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\dots+\frac{1}{f(99)}$
Given $f(n)=\sqrt{2n+1+2\sqrt{n^2+n}}$
Evaluate $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\dots+\frac{1}{f(99)}$
I have simplified the given to $f(n)=\sqrt{n}+\sqrt{n+1}$ but I am still not sure ...
1
vote
1
answer
180
views
Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{3}{2}$
For $a\geq b\geq c >0$. Prove that $$\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}$$
$a=100;b=1;c=1/100$ it's wrong ???
$\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}$ ...
1
vote
2
answers
117
views
Simplifying Multiple Summations for worst case analysis
I'm figuring out a worst case analysis on a function. After converting it to a set of summations, and changing the sigma notations into summation formuale I ended up with:
...
0
votes
6
answers
62
views
How to proof that $\forall n \ge 1: \sum_{i=1}^n (i(i+1))^{-1} = n(n+1)^{-1}$ using mathematical induction
So for a better presentation: https://i.sstatic.net/HCzft.jpg
I need to prove this using mathematical induction, so i started with a base of $n=1$
Since $i$ starts at $i = 1$ I got this:
$$\frac{1}{...
0
votes
2
answers
42
views
Proving Algebraic expression involving a Summation
I'm following a derivation and am stumped by one of the steps:
$\sum_{k=1}^{13}(1 - \frac{k-1}{13})^3 = \frac{1}{13^3}\sum_{k=1}^{13}k^3$
I am stumped as to how
$(14 - (k-1))^3= k^3$
Any help ...
0
votes
3
answers
53
views
How to simplify this expression with fractions?
$$\frac{1}{a(a-b)(a-c)} + \frac{1}{b(b-a)(b-c)} + \frac{1}{c(c-a)(c-b)} $$
I tried to get everything to the same denominator, and then simplify numerators first but it is very complicated and long if ...