Questions tagged [binomial-theorem]
For questions related to the binomial theorem, which describes the algebraic expansion of powers of binomials.
1,791
questions
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How do you find the coefficient of $x$ in $(x + 1)^2$?
I want to learn how can I find out the coefficient of the variable $x$ in the expression $(x + 1)^2$. It is a case of a perfect square expansion.
0
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0answers
18 views
approximated value Using Binomial theorem Expansion
Find An approximated value for
4√630
, (2.9)5
Using (i) Binomial theorem Expansion
(ii) Differentiation Methods
enter image description here
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vote
2answers
62 views
How may one solve problems over expressions like $(2+px)^6$ without the binomial theorem?
A friend of mine posed a problem on a mathematics discord server.
The coefficient of the $x^2$ term in the expansion of $(2+px)^6$ is $60$. Find the value of the positive constant $p$.
I ...
0
votes
1answer
32 views
How do I simplify factorials?
Given the following equation:
$$\frac{n!}{(n-k)!k!}*\frac{1}{n^{k}}=\frac{1}{k!}*\frac{n*(n-1)...(n-k+1)}{n^{k}}$$
(I used * as the multiplication sign).
I understand $\frac{1}{k!}$ but $$\frac{n*(...
0
votes
4answers
52 views
Evaluate the sum $\sum_{n=0}^{\infty} \frac{2n}{8^n}{{2n}\choose{n}}$
Evaluate the sum $\sum_{n=0}^{\infty} \frac{2n}{8^n}{{2n}\choose{n}}$
I am unable to find a way to solve this sum. I have never seen sums involving binomial coefficients multiplied by $n$. Help will ...
0
votes
1answer
51 views
Find the generating function and the number of integers solutions for $x_1 + x_2 + x_3 + x_4 = r$, where $-3 \leq x_i \leq 3$. [closed]
Well, we have $x_1 + x_2 + x_3 + x_4 = r$ where $x_i\in\{-3,-2,-1,0,1,2,3\}$
Then the generating function is given by $f(x)=(x^{-3}+x^{-2}+x^{-1}+x^{0}+x^1+x^2+x^3)^4 = \frac{(x^{6}+x^{5}+x^{4}+x^{3}+...
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votes
3answers
31 views
Closed for solution for $\sum_{k = 0}^{n} Q^{k} ( 1 - Q) ^ {k}$
I know the binomial expansion formula:
$$
(1 + x)^n = \sum_{k = 0}^{n} {n \choose k}x^k
$$
However, I am trying to find (if there is any) a closed-form solution for the following equation.
$$
\sum_{...
0
votes
0answers
39 views
calculate the given sum (related to Newton's binomial formula) [duplicate]
i was given the following question:
given n is a positve integer, calculate the given sum:
$$
{n \choose 1} -2{n \choose 2} +3{n \choose 3}+ \ldots +(-1)^{n-1}\cdot n {n \choose n}
$$
I looked at ...
1
vote
1answer
35 views
Radius of convergence for binomial series (2)
I'm having trouble calculating the radius of convergence for for the following binomial series.
More in detail, I'm having trouble finding $c_k$ and $c_{k+1}$ for the following series:
$$ \sum_{k=0}^...
0
votes
0answers
37 views
Binomial expansion of $(x+1)^{-1}$
Why are the binomial expansions of $(x+1)^{-1}$ and $(1+x)^{-1}$ different?
The expansion of $(1+x)^{-1}$ is -
$$(x+1)^n= 1+ nx + \frac{n(n-1)}{ 2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3\cdots$$
where the ...
0
votes
2answers
40 views
How do I solve this problem using binomial theorem with square root in it?
The question is: Find the coefficient of $\frac{1}{x\sqrt x}$ in the expansion of $(x^2 - \frac{1}{2\sqrt x})^{18}$.
I have included a photo to make it easier to read because I do not know how to ...
0
votes
1answer
45 views
How can I show binomial series converges to $\sqrt {2}$?
$\sum_{n=0}^{\infty} 2n\binom{-\frac{1}{2}}{n}(-\frac{1}{2})^n = \sqrt{2}$
From wolfram alpha, it says that above series including binomial term $\binom{-\frac{1}{2}}{n}$ converges to $\sqrt{2}$.
I ...
2
votes
1answer
50 views
Evaluation of a tricky binomial sum
The Question:
To prove that:
$\frac{3!}{2(n+3)} = \sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}{r}}}$
My Attempt:
I start off by writing $\sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}...
0
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1answer
32 views
Reciprocal of square root of a binomial to series.
by square root algorithm and long division (or by binomial theorem) it is simple matter to find $1/\sqrt{(1-x^2)} = (1+x^2/2 + 3x^4/8 + 5x^6/16 + ...)$ >
I am new to this kind of thing. Can someone ...
0
votes
1answer
44 views
How to show $\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2} = (-1)^m\binom{2m}{m}$ [duplicate]
I am trying to show that for any positive integer m,
$$\sum_{i=0}^{2m} (-1)^{i}\binom{2m}{i}^{2} = (-1)^m\binom{2m}{m}$$
Intuitively this seems to be true, $m = 0$ both sides evaluate to $1$, $m = 1$...
0
votes
0answers
20 views
Inverse of the difference of matrices
Let $A\in R^{n\times n}$ (invertible and symmetric), $B\in R^{k\times k}$ (invertible and symmetric), $C\in R^{n\times k}$, $D\in R^{k\times n}$, $CBD$ is (invertible and symmetric). I am trying to ...
0
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0answers
15 views
Find the misterious function $F(r,a,q)$.
I've the pattern bellow where we can spot the binomial coefficients. Observing the pattern, for each row, it seems that it arises from
$$
\left(\left(r a^{q} \right)^{\frac{1}{q}} +F(r,a,q)\right)^{n}
...
1
vote
1answer
44 views
Sum of product of binomials
While working on a combinatorics problem, I found that this result had to be true:
$$\sum_{i=0}^n\frac{(a-b)^i(b-c)^{n-i}}{i!(n-i)!}=\frac{(a-c)^n}{n!}$$
for $a\geq b\geq c$, with $a,b,c\in\Bbb N$.
...
0
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2answers
50 views
Series Expansion and Big-Oh Notation: Expanding $\sqrt{x^2+\mathcal{O}(x^3)}$ to get $x+\mathcal{O}(x^{3/2})$
I've been reading through a thesis and trying to rederive all of the equations used. However, I've come across an expansion that I'm unsure on. The following is a simplified form of two consecutive ...
1
vote
2answers
34 views
Confused about the meaning behind making x = 2 in the binomial theorem
So the binomial theorem states $(1+x)^n = \sum^{n}_{k=0}$$n \choose k$$x ^ k$.
Now I understand that each term of the sum represents the number of ways to arrange 1 and $x$ out of $n$ choices, so ...
0
votes
1answer
21 views
Radius of convergence for binomial series
The question is simple: determine the radius of convergence for the given series.
I have done this for other problems using the following rule which our book described:
If the given series is
$$ \...
4
votes
1answer
128 views
Tighter upper bound on $x$ where $2^x \leq \sum_{i=0}^m{{x \choose i}\lambda^i}$
We have the following inequality:
$$2^x \leq \sum_{i=0}^m{{x \choose i}\lambda^i}$$
All the variables are in $\mathbb{N}_{>0}$
I need to find a tight upper bound for $x$ using $m,\lambda$.
In ...
1
vote
1answer
58 views
why can't we use binomial theorem to expand this expression
I was recently told that while expanding expressions of the form
$$(A+B)^n$$
where A and B are square matrices of same order and n is a natural number then
$$(A+B)^n =$$
$$ {{n}\choose{0}}A^{0}B^{...
0
votes
1answer
32 views
May I know why is the fact that P(The best is among the first n) is calculated in this way?
I am trying to solve this problem:
On the basis of an interview, the N candidates for admission to a college are ranked in order according to their mathematical potential. The candidates are ...
0
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0answers
19 views
binomial theorem combinatorics consideration question [duplicate]
i have an identity that i need to prove in 2 ways
$$\sum_{k=m}^n \binom k m = \binom {n+1} {m+1}$$
the first way was to prove using induction which wasn't a problem since n≥m so just do the base and ...
0
votes
1answer
21 views
Fractional exponent for binomial theorem
If I am trying to expand $(a+b)^{\frac{2}{3}}$, can I use the binomial theorem like so:
$$\sum_{k=0}^{\frac{2}{3}}{\frac{2}{3}\choose k}a^{\frac{2}{3}-k}b^k$$
or will that not work, since the last ...
0
votes
1answer
59 views
$q$-analogue of $\sum_{k=0}^n \, {n \choose k} = 2^n $
Is there a $q$-analogue of the formula $\sum_{k=0}^n \, {n \choose k} = 2^n $ in terms of the $q$-binomial coefficient ${n \choose k}_q$ and $(2^n)_q=(1+q)...(1+q^n)$?
1
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1answer
40 views
Proving $(1+x)f'(x)=a f(x)$ where $f(x)=\sum_{n=0}^\infty {a\choose n}x^n$
A function $f$ is defined for $-1<x<1$ by $f(x)=\sum_{n=0}^\infty {a\choose n}x^n$. Here $a$ is a real number which is neither zero nor a positive integer. Prove that $(1+x)f'(x)=a f(x)$.
I took ...
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1answer
25 views
Binomial coefficient (calculate m) [closed]
The problem asks: Calculate m knowing that $\left(\begin{matrix}m \\1 \end{matrix}\right) + \left(\begin{matrix}m \\2 \end{matrix}\right) + \left(\begin{matrix}m \\3 \end{matrix}\right)+ \ldots +\left(...
0
votes
1answer
45 views
Use combinatorial reasoning to show that Stirling number
Use combinatorial reasoning to show
$\begin{Bmatrix}
n\\
n-2
\end{Bmatrix} = \binom{n}{3} + 3\binom{n}{4}.$
The Stirling number is the number of permutation of n into $n-2$ parts.
2
votes
1answer
48 views
Calculating $\sum_{k=0}^{n}\binom{n}{k}\frac{1}{k+1}$
Hello everyone how can I calculate the sum of:
$\sum_{k=0}^{n}\binom{n}{k}\frac{1}{k+1}$
I tried to use pascal identity and the binomial theorem and didn't success.
Someone can help me please?
2
votes
0answers
33 views
Proof of $a^n<(a+h)^n<a^n+2nha^{n−1}$ for $h>0$ and sufficiently small $h$
I was reading this answer for the proof of
$$\lim_{x\to a}{x^n}=a^n\ \forall \ n \in \mathbb{R}$$
Now..according to the answer given, the proof uses the following inequality
$$a^n<(a+h)^n<a^n+...
0
votes
0answers
25 views
Using telescoping series to express a partial binomial series
I am analying
$$
\sum_{j=k}^{n}\frac{(-1)^{j}{\frac{-3}{2}\choose n-j}}{j(j-1)}\;\;\mbox{where}\;n\ge k\;\mbox{are integers}.
$$
By the telescoping series, we have
$$
\sum_{j=k}^{n}(-1)^{j}{\frac{-3}{...
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1answer
32 views
How to prove the closed form for the generating function $g_r(x) = \sum_{n=0}^\infty {n \choose r} x^n$ possibly using Newton's Binomial Theorem
define the generating function $g_r(x) = \sum_{n=0}^\infty {n \choose r} x^n$
how do you find the closed form for this function?
I got $x^r{(1-x)}^{-(r+1)}$ but I am not sure how to prove it. I ...
0
votes
2answers
27 views
Binomial coefficient after expansion
I am trying to solve an exercise where in the final step, I need to find the coefficient of $x^7y^5$ in $(x+y)^{12} + 7(x^2+y^2)^6 + 2(x^3+y^3)^4 + 2(x^4+y^4)^3 + 2(x^6+y^6)^2 + 4(x^{12}+y^{12}) + 6(x+...
0
votes
1answer
28 views
Binomial series: why do we take $n \cdot (n-1)$ out?
Given: $\sum_{k=1}^n \binom{n}{k} k(k-1)$. Find its closed form
We can multiply $(k^2-k)$ with $\binom{n}{k}$ and get:
$\frac{n!}{(k-2)!(n-k)!}$
Why do we take $n*(n-1)$ out: $n*(n-1)$*$\frac{(n-2)!...
3
votes
2answers
49 views
Prove combinatorics equality?
Assume $j$ is fixed, prove the following:
$$\sum_{i}\binom{n}{i, j, n-i-j} = 2^{n-j}\binom{n}{j}$$
So the left hand side reminds me the multinomial theorem and we can think of a long sequence word ...
0
votes
3answers
50 views
Show that $n^5-n$ is divisible by 80 given that n is an odd integer greater than 1. [duplicate]
I tried to prove it by simple binomial expansion but I am stuck.
I assumed, $n=(2k+1) ,k=1,2,3,......$ and after expanding the expression $(2k+1)^5-(2k+1)$ I arrived at this expression$→$ $8k+40k^2+...
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0answers
24 views
Proof of binomial theorem for non-integers
Firstly, yes, I know how to expand expressions like $(a+b)^{({1}/{2})}$ but I want a rigorous proof of why it is okay to replace binomial coefficients like ${1/2}\choose{2} $ with $({1/2})({1/2-1}))/2$...
0
votes
1answer
37 views
Prove $\binom{n}{k_1,…,k_m} = \sum_{i=1}^m \binom{n - 1}{k_1,…,k_{i - 1},..,k_m}$
I have a question ask to prove $$\binom{n}{k_1,...,k_m} = \sum_{i= 1}^m \binom{n - 1}{k_1,...,k_{i - 1},..,k_m}$$
I'm not sure how to approach this question, but the only thing that I noticed the LHS ...
6
votes
5answers
175 views
Factorise $x^4+y^4+(x+y)^4$
I think this problem involves making use of symmetry in some way but I don't know how. I expanded the $(x+y)^4$ term but it din't help in the factorization. I am very bad at factorizing, so I didn't ...
0
votes
4answers
42 views
Find the coefficient of $x^{-2}$ in the expansion of $(x-1)^3(\frac{1}{x} +2x)^6$
Find the coefficient of $x^{-2}$ in the expansion of $(x-1)^3(\frac{1}{x} +2x)^6$.
I have tried finding out the general term of the whole expression.
Then, I get the result as such
$$[(-1)^r {3 \...
-1
votes
1answer
26 views
$\sum_{r=0}^{n-k} {n \choose r}{n \choose {r+k}}={n \choose {n-k}}$ [duplicate]
$$\sum_{r=0}^{n-k} {n \choose r}{n \choose {r+k}}={n \choose {n-k}}$$
Can you prove why this result is as such? I have no idea how to start proving this property.
Thank You!
0
votes
0answers
51 views
Determining the minimum score to pass
I am revising some basic statistics problems for my studies and came across this problem:
A population of 10 000 students is taking an exam and the exam score is follows approximately a normal ...
3
votes
1answer
82 views
A Conceptual Intepretation of the identity
I came across the identity
$$\sum_{k = 0}^{n}\frac{(-1)^{k}\binom{n}{k}}{2k+1} = \frac{2^{2n}(n!)^2}{(2n+1)!}$$
I tried it using the binomial theorem and integration and was able to prove it provided ...
1
vote
3answers
40 views
Sum of a binomial sequence equation
Anyone know how to go about solving this? I've tried adding the individual terms with no luck, and also I don't get how the 'n' in 15Cn doesn't correspond to (-1)^n ...
-1
votes
1answer
54 views
Is there a formula for $(a+b)^n+(a-b)^n$?
Is there a formula for $(a+b)^n+(a-b)^n$? Just curious. Thanks.
0
votes
5answers
63 views
proof $k!j!\leq(j+k)!$ by induction [duplicate]
how do I prove that $k!j!\leq(j+k)!$
I have tried to use induction but didn't succeed.
any other ideas on how to prove it?
in witch case $k!j!=(j+k)!$ ?
5
votes
2answers
90 views
Proving interesting identity with partial sums of Pascal's triangle rows
As part of another problem I'm working on, I find myself needing to prove the following:
$$\sum_{k=0}^n\binom{2k+1}{k}\binom{m-(2k+1)}{n-k} = \sum_{k=0}^{n}\binom{m+1}{k}$$
where $n\leq m$. I've ...
2
votes
1answer
30 views
Find coefficients of $x^2$ and $x^3$ in the expansion of $(3 − 2x) ^6$
The coefficients of $x^2$ and $x^3$ in the expansion of $(3 − 2x) ^6$ are $a$ and $b$ respectively. What is the value of $a$ and $b$?
They need to be good at the binomial theorem and know the ...
