Tagged Questions
4
votes
1answer
42 views
Binomial theorem for prime exponent
Could you explain to me why for prime $p$ we have the following?
$$(x+y)^p - (x^p + y^p)= x^p + \binom{p}{1}x^{p-1}y + \binom{p}{2}x^{p-2}y^2 + \binom{p}{p-1}xy^{p-1} + y^p.$$
I found it here: ...
0
votes
2answers
45 views
Calculation of binomial sum $\displaystyle \sum_{r=1}^{n}r.\binom{n}{r}x^r.(1-x)^{n-r} = \;\;?$
How can I calculate
$$\displaystyle \sum_{r=1}^{n} r \binom{n}{r}x^r (1-x)^{n-r} =\;\; ?$$
5
votes
4answers
107 views
Proving $\binom{2n}{n}\le 4^n$ for all $n$ by smallest counterexample
Prove $$\binom{2n}{n}\le 4^n$$ for all natural numbers $n$ by smallest (minimal) counterexample.
My attempt:
First, $$\binom{2n}n = \frac{(2n)!}{(n!)^2} \le 4^n\;.$$ We know that $x\ne 0$ because ...
6
votes
3answers
137 views
How many solutions are possible to this equation?
Given
$$A+2B+3C=N
$$
where $N$ is a given positive integer.
$A ,B,C\in\mathbb{N}$ vary from $0$ to $\infty$.
How many solutions will be there to this equation?
6
votes
1answer
127 views
Closed-form expression for $\sum_{n=1}^{k} (-1)^{n+1}n^2(n^2-1)\binom{2k}{k-n}$?
Wolframalpha tells me that
$$\sum_{n=1}^{k} (-1)^{n+1}n^2(n^2-1)\binom{2k}{k-n}=0$$ for $k>2$
However I have not been able to come up with a proof and I simply don't see how to do it. Does anyone ...
1
vote
3answers
109 views
Can the identity $n(n+1)2^{n-2} = \sum_{i=1}^{n} i^2 \binom{n}{i}$ be derived from the binomial theorem?
Can this identity be derived from the binomial theorem?
$n(n+1)2^{n-2} = \sum_{i=1}^{n} i^2 \binom{n}{i}$
Please, explain how.
I tried starting from $2^n = \sum_{i=0}^{n} \binom{n}{i}$ and ...
0
votes
1answer
53 views
How did my textbook conclude this proof?
http://i.imgur.com/goUlA.png
At the very last step, highlighted in red, it states that if this, then that sort of, and I'm not sure how those comparisons explain the issue of proving Pascal's ...
0
votes
1answer
64 views
Could anyone explain how my textbook did this simplification?
The book is talking about proving Pascal's triangle increases until the middle, until which point it decreases.
Theorem 4.2 refers to the Multiplicative formula
I just don't understand how that ...
0
votes
2answers
322 views
Calculate reciprocal square root via binomial expansion
I have to use a binomial expansion to evaluate $1/\sqrt{4.2}$ to $5$ decimal places. The answer from a calculator is $0.48795$ but I get $0.48202$, so I'm doing something wrong. I've also checked my ...
2
votes
2answers
482 views
Find the coefficient of $x^3y^2z^3$ in the expansion $(2x+3y-4z+w)^9$
The exercise says:
In the expansion $(2x+3y-4z+w)^9$, find the coefficient of
$x^3y^2z^3$.
The formula to find the coefficient of $x_1^{r_1}x_2^{r^2}\dots x_k^{r_k}$ in $(x_1+x_2+\dots+x_k)^n$ ...
2
votes
2answers
249 views
Does this qualify as a proof? (Spivak's 'Calculus')
I'm working through Spivak's 'Calculus' at the moment, and a question about series confused me a bit. I think I have the solution, but I'm not sure if my "proof" holds.
The question is:
Prove ...
4
votes
0answers
87 views
How to transform series of series into series
I need to prove this equation.
$$ \sum_{k=0}^{i-2} \left( e \space α(k+1)\space\frac{(-1)^{i+k+2}}{(i-k-2)!} \right) = \sum_{k=0}^{i-2} \frac{(i-k)^k}{k!} \space e^{i-k} (-1)^k\space where,\space α(i) ...
1
vote
2answers
161 views
simplify $(a_1 + a_2 +a_3+… +a_n)^m$
How to simplify this best
$(a_1 + a_2 +a_3+... +a_n)^m$
for
$m=n, m<n, m>n$
I could only get
$\sum_{i=0}^{m}\binom{m}{i}a_i^i\sum_{j=0}^{m-i}\binom{m-i}{j}a_j ... $
2
votes
2answers
107 views
Why is the binomial coefficient related to the binomial theorem?
The binomial coefficient basically provides the number of ways to choose a set of $k$ from $n$ sets. To me, it can be considered the number of unique ways to pick $k$ amount of "cards" from a deck of ...
6
votes
4answers
254 views
Showing that $\lceil (\sqrt{3} + 1)^{2n} \rceil$ is divisible by $2^{n+1}$.
I have a question which has fluxommed me and my pals for the past few days. Any help or solution is welcome
Show using Binomial theorem that the integer just after $(3^{1/2} + 1)^{2n}$ is divisble ...
4
votes
5answers
203 views
The sum of the coefficients of $x^3$ in $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$
I know how to solve such questions when it's like $(x+y)^n$ but I'm not sure about this one:
In $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$, What's the sum of the
coefficients of $x^3$?
7
votes
3answers
391 views
Hard elementary combinatorics problem
How does one compute (without brute force) the smallest integer $n$ such that
$\binom{2n}{1}(-3)^0 + \binom{2n}{3}(-3)^1 + \binom{2n}{5}(-3)^2 + \cdots + \binom{2n}{2n-1}(-3)^{(n-1)} = 0$?
3
votes
5answers
251 views
Binomial theorem application
I have a question about the bonomial theorem, and in specifically, a question that I want help on. I have worked out the answer, but by manually expanding each and every alternative. However, I ...
1
vote
3answers
482 views
Calculate the expansion of $(x+y+z)^n$
The question that I have to solve is an answer on the question "How many terms are in the expansion?".
Depending on how you define "term" you can become two different formulas to calculate the terms ...
2
votes
3answers
273 views
Water and wine mixing problem
This is a well-known problem involving a water barrel and a wine barrel, described here. The trick to solving the puzzle is that one need not make the calculations for each stage of the liquid ...
12
votes
5answers
367 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!$$
...
4
votes
2answers
309 views
How can I compute $\sum\limits_{k = 1}^n \frac{1} {k + 1}\binom{n}{k} $?
This sum is difficult. How can I compute it, without using calculus?
$$\sum_{k = 1}^n \frac1{k + 1}\binom{n}{k}$$
If someone can explain some technique to do it, I'd appreciate it.
Or advice using ...
1
vote
1answer
77 views
A “fast” approach to compute $\sum_{i=0}^{n} \binom{19}{i} \times \binom{7}{n-i}$ [duplicate]
Possible Duplicate:
How to find a closed formula for the given summation
I am looking for a fast/best approach to compute $$\sum_{i=0}^{n} \binom{19}{i} \times \binom{7}{n-i}$$
For ...
1
vote
1answer
94 views
Combination Problem with a Variable
I have the following problem:
$_xC_6$ = $_xC_4$
I expand both sides to:
$$\frac{x!}{[(x-6)!]6!} = \frac{x!}{[(x-4)]!4!}$$
Next I multiply both sides by the denominator of the right-hand ...
3
votes
3answers
263 views
What does the notation $\binom{n}{i}$ mean?
What do the parentheses next to the summation involving the binomial coefficients mean? Like this:
$$\sum _{i=0}^{n} \binom{n}{i}a^{(n-i)}b^i=\left(a+b\right)^n $$
2
votes
3answers
91 views
Further simplify this $\sum \limits_{i=1}^{k}{{k \choose i} \cdot 12^i \cdot 2^i}$
Could we further simplify this:
$$ \sum_{i=1}^{k}{{k \choose i} \cdot 12^i \cdot 2^i}$$
or, at least, find a close upper bound?
12
votes
4answers
393 views
How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$?
Other than the general inductive method,how could we show that $$\sum_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$$
Apart from induction,I tried with Wolfram Alpha to check the validity,but ...
3
votes
1answer
127 views
Simplification of a sum
Any ideas on how to approximate and/or simplify this crazy-looking sum will be massively appreciated)
$$\frac{1}{\mu}\sum_{j=0}^{\frac{\lambda}{2}} ...
2
votes
3answers
147 views
How can I prove the formula for calculating successive entries in a given row of Pascal's triangle?
I've found in Wikipedia the formula for calculating an individual row in Pascal's Triangle:
$$v_c = v_{c-1}\left(\frac{r-c}{c}\right).$$
where $r = \mathrm{row}+1$, $c$ is the column starting from $0$ ...
10
votes
3answers
679 views
Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$
I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subsets of a set of $n$ elements. I'm curious if there's a ...
1
vote
1answer
126 views
How can we prove this property?
$ \qquad \qquad $ The greatest term in $\left(1+x\right)^{2n}$ has the greatest cofficient if $\frac{n}{n+1} \lt x \lt \frac{n+1}{n}$
Can we derive something like this for $n$ in general? Please ...
2
votes
4answers
157 views
How to sum up this series?
How to sum up this series :
$$2C_o + \frac{2^2}{2}C_1 + \frac{2^3}{3}C_2 + \cdots + \frac{2^{n+1}}{n+1}C_n$$
Any hint that will lead me to the correct solution will be highly appreciated.
EDIT: ...
2
votes
2answers
90 views
Are these two expressions for sums of binomial coefficients valid?
These two appear in my module (without any proof):
$$\sum_{r = 1}^{n} C(n-2,r-2) = 2^{n-1}$$
$$\sum_{r = 0}^{n-1} C(n-2,r) = 2^{n-2}$$
For the first one when when $r=1 \Rightarrow C(n-2,1-2) = ...
8
votes
3answers
374 views
Non-probabilistic proofs of a binomial coefficient identity from a probability question
Combining the answers given by me and Ralth to the probability question at Probability Question, we get the following identity:
$$
\sum\limits_{k = m}^n {{n \choose k}p^k (1 - p)^{n - k} {k \choose ...
4
votes
1answer
114 views
How to calculate this efficiently?
If in the expansion of $(1 + x)^m \cdot (1 – x)^n $, the coefficients of $ x $ and $ x^2 $are 3 and -6 respectively, then m is ?
I solved it in the following way :
Expanding we get, the coefficient ...
1
vote
4answers
328 views
Showing $\frac {1}{(n-1)!} + \frac {1}{3!(n-3)!} + \frac {1}{5!(n-5)!} +\frac {1}{7!(n-7)!} + \cdots = \frac {2^{n-1}}{(n)!} $
I am stuck while proving this identity I verified it using induction but like the other two (1) (2), I am seeking a more of a general way (algebraic will be much appreciated)
$$\frac {1}{(n-1)!} + ...
3
votes
5answers
3k views
Proving $ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $
How to prove this binomial identity :
$$ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
The left hand side arises while solving a standard binomial problem the right hand ...
5
votes
10answers
1k views
How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$?
I am trying to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here, so I am ...
