Binomial theorem Definition and 136 Threads

This answer shows an extended version of Pascal's Triangle that works for negative numbers too. In This video, Sal shows how to interpret the members of Pascal's Triangle as the sum of all the possible paths to get to that member. Is there any way we can use this same 'sum of all the possible...

##r^{th}## term from beginning: ##^nC_{r-1}a^{n-r+1}(2x)^{r-1}## For the ##r^{th}## term from the end, we first know there are a total of ##n+1## terms in this binomial expansion. Subtracting the (##r^{th}## term from the end) from the total number of terms, ##n+1##, results in ##n+1-r## which...

The book solution is to first take one 5 out 5(5^98)= 5(25^49)=5(26-1)^49 And then when we expand it using Binomial theorem we get a number which isnt a multiple of 13, we get -5 as the remainder. But since remainders have to be positive we add 13 to it (this i generalised by dividing numbers...

##r^{th}## term counting from the beginning: The coefficient of the ##r^{th}## term is ##r-1## ##^nC_{r-1}a^{n-(r-1)}(2x)^{r-1} = ^nC_{r-1}a^{n-r+1}(2x)^{r-1}## This is the correct answer. ##r^{th}## term counting from the end: There are a total of ##n+1## terms in ##(a+2x)^n##...

By definition, ##^nC_r=\dfrac{n(n-1)(n-2)...(n-r+1)}{r!}##. This can be simplified to ##^nC_r=\dfrac{n!}{r!(n-r)!}##, which leads to ##^{2m}C_m=\dfrac{2m!}{m!m!}##. But I can't see how from the original equation ##^{2m}C_m=\dfrac{(2m)(2m-1)(2m-2)...(m+1)}{m!}## is equivalent to...

I'm sort of stumped here , do i do this? (1+3x) \left( \frac{1+3x}{1+2x} \right)^2 = (1+3x) \left( \frac{3}{2} - \frac{1}{2(2x+1)} \right)^2 (1+3x) \left( \frac{3}{2} \right)^2 \left( 1 + \frac{-1}{3(2x+1)} \right)^2 and then apply the binomial theorem formula on the squared term above...

part (a) ##(4+3x)^{1.5} = 2^3+ 9x+ \left[\dfrac {1}{2} ⋅ \dfrac {3}{2} ⋅\dfrac {1}{2}⋅\dfrac {1}{2}⋅9x^2\right]+ ...## ##(4+3x)^{1.5}=8+9x+\dfrac {27}{16} x^2+...##part (b) ##x≠-\dfrac {4}{3}##part (c) ##(8+9x+\dfrac {27}{16} x^2+...)(1+ax)^2 = \dfrac{107}{16} x^2## ... ##8a^2+18a+\dfrac...

This is also pretty easy, ##z^5=(a+bi)^5## ##(a+bi)^5= a^5+\dfrac {5a^4bi}{1!}+\dfrac {20a^3(bi)^2}{2!}+\dfrac {60a^2(bi)^3}{3!}+\dfrac {120a(bi)^4}{4!}+\dfrac {120(bi)^5}{5!}## ##(a+bi)^5=a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i## ##\bigl(\Re (z))=a^5-10a^3b^2+5ab^4## ##\bigl(\Im (z))=...

$$(7-6x)^3+(7+6x)^3=1736$$ $$⇒(7^3(1-\frac {6}{7}x)^3+(7^3(1+\frac {6}{7}x)^3=1736$$ $$343[1-\frac {18}{7}x+\frac {216}{98}x^2-\frac{1296}{2058}x^3]+343[1+\frac {18}{7}x+\frac {216}{98}x^2+\frac{1296}{2058}x^3]=1736$$ $$343[2+\frac {432}{98}x^2]=1736$$ $$686+\frac {148,176}{98}x^2=1736$$ $$\frac...

In the general expansion of (1+x)^n what does the sum of the coefficients mean?

Hi. Is the binomial theorem ##(1+x)^n = 1+nx+(n(n-1)/2)x^2 + ….## valid for x replaced by an infinite series such as ##x+x^2+x^3+...## with every x in the formula replaced by the infinite series ? If so , does the modulus of the infinite series have to be less than one for the series to...

Determine an expression for f(x) =(1+x)(1+2x)(1+3x)…(1 +nx),find f⸍(0) .

I'm trying to expand the following using Newton's Generalized Binomial Theorem. $$[f_1(x)+f_2(x)]^\delta = (f_1(x))^\delta + \delta (f_1(x))^{\delta-1}f_2(x) + \frac{\delta(\delta-1)}{2!}(f_1(x))^{\delta-2}(f_2(x))^2 + ...$$ where $$0<\delta<<1$$ But the condition for this formula is that...

Homework Statement The sixth term of the expansion of (x-1/5)n is -1287/(3125)x8. Determine n. Homework Equations tk+1=nCkan-kbk The Attempt at a Solution tk+1=nCkan-kbk t5+1=nC5(x)n-5(-1/5)5 This is where I'm stuck. Do I sub in -1287/(3125)x8 to = t6? If so what do I do from here...

Homework Statement 1. Given the binomial (x2-x)13determine the coefficient of the term of degree 17. Answer = -715 2. Given the binomial (2x+3)10 determine the coefficient of the term containing x7. Answer = 414720 2. Homework Equations tk+1=nCkan-kbk The Attempt at a Solution #1 - What...

Hello! When studying binominal expansion: ## (a+b)^n = \sum_{k=0}^{n}{{n \choose k}a^{n-k}b^k} ## in high school, we proved this formula with combinatorics considering that "you can choose either a or b each time you multiply with a binom". Probably, this is not a real mathematical proof at...

Homework Statement Note: I'm saying it's very very hard because I still couldn't solve it and I've posted it in stackexchange and no answer till now. I'm posting here the problem statement, all variables and known data in addition to my solving attempts. Because I'm posting an image of my...

Homework Statement Expand x(k+1)/(k+1) - (x-1)(k+1)/(k+1) Homework Equations (a+b)m = am + mam - 1b + (mℂ2)am - 2b2 + ... + bm[/B] The Attempt at a Solution Here is my solution, I would like to know if it's correct or not I have the solution in an attached image

I uploaded a picture of what I am stuck on. I understand the equation of motion 3.4.5a for a damped oscillator but I don't understand how to use binomial theorem to get the expanded equation 3.4.5b. I am no where near clever enough to figure this one out. I know how to use binomial theorem to...

Homework Statement Using the Generating function for Legendre polynomials, show that: ##P_n(0)=\begin{cases}0 & n \ is \ odd\\\frac{(-1)^n (2n)!}{2^{2n} (n!)^2} & n \ is \ even\end{cases}## Homework Equations Generating function: ##(1-2xt+t^2)^{-1/2}=\displaystyle\sum\limits_{n=0}^\infty...

Simplify (find the sum) of $${30 \choose 0} + \frac{1}{2}{30 \choose 1}+ \frac{1}{3}{30 \choose 2} + ... + \frac{1}{31}{30 \choose 30}$$. Do this is two ways: 1. Write $$\frac{1}{i+1}{30 \choose i}$$ in a different way then add 2. Integrate the binomial thorem (don't forget the constant of...

Homework Statement Find the Laurent Series of f(z) = \frac{1}{z(z-2)^3} about the singularities z=0 and z=2 (separately). Verify z=0 is a pole of order 1, and z=2 is a pole of order 3. Find residue of f(z) at each pole. Homework Equations The solution starts by parentheses in the form (1 -...

Homework Statement Prove that \sum\limits_{k=0}^l{n \choose k}{m \choose l-k} = {n+m \choose k}Homework Equations Binomial theorem The Attempt at a Solution [/B] We know that (1+x)^n(1+x)^m = (1+x)^{n+m} which, by the binomial theorem, is equivalent to: {\sum\limits_{k=0}^n{n \choose...

Homework Statement Find \sum\limits_{k=0}^{n}k^2{n\choose k}(\frac{1}{3})^k(\frac{2}{3})^{n-k} Homework Equations -Binomial theorem The Attempt at a Solution I am using the binomial coefficient identity {n\choose k}=\frac{n}{k}{{n-1}\choose {k-1}}: \sum\limits_{k=0}^{n}k^2{n\choose...

So I was checking the How to self-study math thread and saw that someone suggested that It would be helpfull to create this kind of thread. And because we are writting a test on thursday on Probability I though it would be nice to find out which parts I still need to double-check. So these...

Homework Statement How to use binomial theorem for finding sums with binomial coefficients? Example: S={n\choose 1}-3{n\choose 3}+9{n\choose 5}-... How to represent this sum using \sum\limits notation (with binomial theorem)? Homework Equations (a+b)^n=\sum\limits_{k=0}^{n}{n\choose...

Homework Statement For reference, this is the image setting up the problem. "A wireless sensor grid consists of 21×11=231 sensor nodes that are located at points (i,j) in the plane such that i∈{0,1,⋯,20} and j∈{0,1,2,⋯,10} as shown in Figure 2.1. The sensor node located at point (0,0) needs...

Homework Statement Prove that ∑nj=0(-1)j(nCj)=0Homework Equations Definition of binomial theorem. The Attempt at a Solution If n∈ℕ and 0≤ j < n then 0=∑nj=0(-1)j(nCj) We know that if a,b∈ℝ and n∈ℕ then (a+b)n=∑nj=0(nCj)(an-jbj) Let a=1 and b= -1 so that 0=(1+(-1))n=∑nj=0(nCj)(1n-j(-1)j)...

The below image shows a portion of my current Analytical Mechanics textbook. My inquiry is how is the binomial theorem used to get from eq. 3.4.5a ⇒ 3.4.5b ? Thanks in advance

Homework Statement ##\sum\limits_{r=0}^n\frac{1}{^nC_r}=a##. Then find the value of $$\sum\sum\limits_{0\le i<j\le n}(\frac{i}{^nC_i}+\frac{j}{^nC_j})$$ Homework Equations I have used two equations which I derived myself. This is the first one. The second one is: 3. The Attempt at a...

Any hints for this : 1- (11C1/2.3 ).2^2 + (11C2/3.4 ). 2^3 ...so on up to 12 terms .

Screenshot by Lightshot The translation in binom coefficent of 4th and 10th are mathching each other. Find the member which doesn't have x in it. I understand all of it but the part where (n up n-3)=(n up 9) I just don't understand how they got 12 here

In solved binominal form (4x+3)^n has two members x^4 and x^3 whose binomial coefficients are equal. I'm kinda good in solving binomial coefficient, but I never stumbled to something like this

Homework Statement From an old exam: Show that \begin{equation*} \sum_{0 \leq 2k \leq n} \binom{n}{2k}2^k = 0 (3) \text{ iff } n = 2 (4). \end{equation*} By ##a = b (k)## I mean that ##a## is congruent to ##b## modulo ##k##. Homework Equations Binomial theorem: ## (a + b)^m =...

[SIZE="4"]Definition/Summary The binomial theorem gives the expansion of a binomial (x+y)^n as a summation of terms. The binomial theorem for positive integral values of 'n', is closely related to Pascal's triangle. [SIZE="4"]Equations The theorem states, for any n \; \epsilon \...

From the binomial theorem, we have $\displaystyle \begin{align*}\left(1+\dfrac{1}{5}\right)^{1000}&={1000 \choose 0}\left(\dfrac{1}{5}\right)^{0}+{1000 \choose 1}\left(\dfrac{1}{5}\right)^{1}+{1000 \choose 2}\left(\dfrac{1}{5}\right)^{2}+\cdots+{1000 \choose...

1. How do you get n!/(k-1)!(n-k+1)! from \begin{pmatrix} n\\k-1 \end{pmatrix} I thought it would be n!/(k-1)!(n-k-1)! where the n-k+1 on the bottom of the fraction would be a n-k-1 instead. I don't understand why there is a "+1" wouldn't you just replace k with k-1 in the binomial formula?

Hello all! This isn't a problem in particular I'm having trouble with, but a much more general question about the binomial theorem. I'm using Stewart's precal book. The section devoted to the theorem has several problems dealing with proving different aspects of it, mostly having to do with...

I know Binomial Theorem is a quick way of expanding a Binomial Expression that has been raised to some power i.e ##(a+b)^n##. But why is it so important to expand ##(a+b)^n##. What is the practical use of this in Science and Engineering.

Here is the question: I have posted a link there to this thread to the OP can view my work.

Hello, I have a problem regarding the binomial theorem and a number of questions about what I can and can't do. Homework Statement Write the binomial expansion of (1 + x)^{2}(1 - 5x)^{14} as a series of powers of x as far as the term in x^{2} Homework Equations The Attempt at a Solution I...

Here is the question: I have posted a link there to this thread so the OP can view my work.

I'm sure this is easy but it has got me baffled. I'm told that the binomial theorem can be used to simplify the following formula x = \dfrac{1 - ay/2}{\sqrt{1-ay}} to (approximately) x = 1 + a^2 y^2 / 8 if a << 1. Thanks for any help or pointers on this one in particular, and/or general...

If it's possible to relate the product rule with the binomial theorem, so: (x+y)^2=1x^2y^0+2x^1y^1+1x^0y^2 D^2(fg)=1f^{(2)}g^{(0)}+2f^{(1)}g^{(1)}+1f^{(0)}g^{(2)} So, is it possible to relate the quotient rule with the binomial theorem too?

Question: If $\displaystyle \sum_{r=0}^{2n} a_r(x-2)^r=\sum_{r=0}^{2n} b_r(x-3)^r$ and $a_k=1$ for all $k \geq n$, then show that $b_n={}^{2n+1}C_{n+1}$. Attempt: I haven't been able to make any useful attempt on this one. I could rewrite it to: $$\sum_{r=0}^{n-1} a_r(x-2)^r +...

On my problem sheet I got asked to prove: ## (1+x)^n = \displaystyle\sum _{k=0} ^n \binom{n}{k} x^k ## here is my attempt by induction... n = 0 LHS## (1+x)^0 = 1 ## RHS:## \displaystyle \sum_{k=0} ^0 \binom{0}{k} x^k = \binom{0}{0}x^0 = 1\times 1 = 1 ## LHS = RHS hence true for...

find the term with $x^2$ $\displaystyle\left(x^2-\frac{1}{x}\right)^{10}$ thanks!

Homework Statement lim x->1 (X^9 + x -2)/(x^4 + x -2) I know how to do this using L'Hopitals Rule and I get 2 Homework Equations (1+b)^n = 1 + bn + n(n-1)b^2/2! + n(n-1)(n-2)b^3/3! ... The Attempt at a Solution Let x = h+1 x -> 1 h -> 0 lim h->0 (h+1)^9 +...

Here is the question: I have posted a link there to this topic so the OP can see my work.

I highlighted the portion in red in the paint document that I'm not understanding. How can we see by inspection that the product is equal to the series 2?