All Questions
Tagged with summation algebra-precalculus
165
questions
16
votes
6
answers
596
views
If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$ [duplicate]
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$
I came across this question in ...
- 6,560
15
votes
1
answer
31k
views
Find the sum $\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{99}+\sqrt{100}}$
I would like to check I have this correct
Find the sum
$$\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{99}+\sqrt{100}}$$
Hint: rationalise the denominators to ...
- 1,155
13
votes
1
answer
2k
views
Notation: What is the scope of a sum?
I would interpret $\sum_{i=1}^2 x_i + y$ as $x_1 + x_2 + y$, but I would interpret $\sum_{i=1}^2 x_i + y_i$ as $x_1 + y_1 + x_2 + y_2$. I realize this is a little inconsistent. Should the latter be ...
- 233
12
votes
5
answers
720
views
How to prove that $\sum\limits_{i=0}^p (-1)^{p-i} {p \choose i} i^j$ is $0$ for $j < p$ and $p!$ for $j = p$
Let $p \in \mathbf{N}$. I don't know how to prove that
$$\sum_{i=0}^p (-1)^{p-i} {p \choose i} i^j=0 \textrm{ for } j \in \{0,\ldots,p-1\},$$
and
$$\sum_{i=0}^p (-1)^{p-i} {p \choose i} i^p=p!$$
(...
- 4,462
12
votes
2
answers
937
views
$\sum\limits_{i=1}^n \frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$, for all $x_i>0.$
Can you prove the following new inequality? I found it experimentally.
Prove that, for all $x_1,x_2,\ldots,x_n>0$, it holds that
$$\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits _{j=...
- 2,040
10
votes
4
answers
418
views
Evaluating $ \sum\limits_{n=1}^\infty \frac{1}{n^2 2^n} $
Evaluate
$$ \sum_{n=1}^\infty \dfrac{1}{n^2 2^n}. $$
I have tried using the Maclaurin series of $2^{-n}$ but it further complicated the question. Moreover, I have also tried taking help from another ...
user208998
9
votes
3
answers
369
views
How to evaluate the sum : $\sum_{k=1}^{n} \frac{k}{k^4+1/4}$
I have been trying to figure out how to evaluate the following sum:
$$S_n=\sum_{k=1}^{n} \frac{k}{k^4+1/4}$$
In the problem, the value of $S_{10}$ was given as $\frac{220}{221}$.
I have tried ...
- 671
7
votes
2
answers
191
views
Sequential sums $1+2+\cdots+N$ that are squares [duplicate]
While playing with sums $S_n = 1+\cdots+n$ of integers,
I have just come across some "mathematical magic"
I have no explanation and no proof for.
Maybe you can give me some comments on this:
I had ...
- 71
7
votes
6
answers
337
views
Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-...
- 4,804
6
votes
4
answers
5k
views
$\sum r(r+1)(r+2)(r+3)$ is equal to?
$$\sum r(r+1)(r+2)(r+3)$$ is equal to?
Here, $r$ varies from $1$ to $n$
I am having difficulty in solving questions involving such telescoping series. While I am easily able to do questions where a ...
- 189
6
votes
6
answers
708
views
Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum
Is there a formula for the following sum?
$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots + n\cdot (n+1)$
- 171
5
votes
2
answers
333
views
Sum this series $\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\ldots$ upto $n$ terms [duplicate]
Sum this series: $$\dfrac{1}{1+1^2+1^4}+\dfrac{2}{1+2^2+2^4}+\ldots$$ upto $n$ terms.
My approach:
$$(1-n^6)=(1-n^2)(1+n^2+n^4)\implies \dfrac{n}{1+n^2+n^4}=\dfrac{n(1-n^2)}{1-n^6}$$
So, the ...
- 6,580
5
votes
3
answers
450
views
Non-induction proof of $2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1$
Prove that $$2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1.$$
After playing around with the sum, I couldn't get anywhere so I proved inequalities by induction. I'm however ...
- 2,916
5
votes
2
answers
601
views
A "fast" way for computing $\sum\limits_{n=1}^{100} n\times 2^n $ [duplicate]
How to compute 'z', where $\displaystyle z = \sum_{n=1}^{100} n\times 2^n$ ?
The answer is of the form $99 \times 2^{101} + 2$, I need a fast approach as this problem is supposed to be solved under a ...
- 22.5k
5
votes
2
answers
2k
views
Sum of derangements and binomial coefficients
I'm trying to find the closed form for the following formula
$$\sum_{i=0}^n {n \choose i} D(i)$$
where $D(i)$ is the number of derangement for $i$ elements. A derangement is a permutation in which ...
- 129