# What is Binomial theorem: Definition and 138 Discussions

In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial (x + y)n into a sum involving terms of the form axbyc, where the exponents b and c are nonnegative integers with b + c = n, and the coefficient a of each term is a specific positive integer depending on n and b. For example (for n = 4),

(
x
+
y

)

4

=

x

4

+
4

x

3

y
+
6

x

2

y

2

+
4
x

y

3

+

y

4

.

{\displaystyle (x+y)^{4}=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}.}
The coefficient a in the term of axbyc is known as the binomial coefficient

(

n
b

)

{\displaystyle {\tbinom {n}{b}}}
or

(

n
c

)

{\displaystyle {\tbinom {n}{c}}}
(the two have the same value). These coefficients for varying n and b can be arranged to form Pascal's triangle. These numbers also arise in combinatorics, where

(

n
b

)

{\displaystyle {\tbinom {n}{b}}}
gives the number of different combinations of b elements that can be chosen from an n-element set. Therefore

(

n
b

)

{\displaystyle {\tbinom {n}{b}}}
is often pronounced as "n choose b".

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1. ### B How to interpret Pascal's Triangle for negative numbers?

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...
2. ### Find the ##r^{th}## term of ##(a+2x)^n##

##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...
3. ### Divisibility and Remainder

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...
4. ### Find the ##r^{th}## term from beginning and end of ##(a+2x)^n##

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5. ### Why is ##^{2m}C_m## equivalent to ##\dfrac{2m!}{m!m!}##?

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...
6. ### Binomial Theorem - determine the term with...

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...
7. ### Solve the given problem that involves binomial theorem

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...
8. ### Use binomial theorem to find the complex number

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))=...

46. ### Binomial theorem proof by induction

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...
47. ### MHB Find Term with $x^2$ in Binomial Theorem

find the term with $x^2$ $\displaystyle\left(x^2-\frac{1}{x}\right)^{10}$ thanks!
48. ### Binomial theorem to evaluate limits?

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 +...
49. ### MHB Apply Binomial Theorem: Expand (x-2y)^3

Here is the question: I have posted a link there to this topic so the OP can see my work.
50. ### What is the solution to the Binomial Theorem problem highlighted in red?

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?