What is the solution to the Binomial Theorem problem highlighted in red?

AI Thread Summary
The discussion centers around understanding a specific aspect of the Binomial Theorem related to a highlighted portion in a document. The participants explore how the term independent of x in the product relates to the series equal to 2. It is clarified that this term is derived from a sum of products involving coefficients and powers of x. A comparison is made between coefficients from two different expressions, revealing they are not equal. Ultimately, the confusion is resolved as one participant acknowledges their understanding of the sum of terms involved.
Miike012
Messages
1,009
Reaction score
0
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?
 

Attachments

  • series.jpg
    series.jpg
    22.4 KB · Views: 613
Physics news on Phys.org
Miike012 said:
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?

The term independent of x in the product is equal to the series (2); this is because the term independent of x is a sum of terms of the form kc_k x^k \times c_k/x^k = kc_k^2.
 
pasmith said:
The term independent of x in the product is equal to the series (2); this is because the term independent of x is a sum of terms of the form kc_k x^k \times c_k/x^k = kc_k^2.
The second line in the paint document is equal to n/xn(1 + (2n-1)x + (2n-1)(2n-1)/2!x2 + ...) by using the binomial theorem

If we look at the coefficient of the second term it is equal to n(2n-1).

If we compare the coeff. n(2n -1) with the coeff. of the second term of series (2) which is
2c22 = 2(n-1)(n-2)/2! = n2 - 3n + 2 they are not equal.

Hence n(2n-1) =/= n2 - 3n + 2
 

Attachments

  • L.jpg
    L.jpg
    5.5 KB · Views: 401
Never mind I see that u said its the sum of the terms... I got it now
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
View full post »

Similar threads

Binomial Theorem - determine the term with...
Replies
18
Views
2K
Why Does Expanding (13-8)^99 Not Yield the Same Remainder as 5^99 Modulo 13?
Replies
7
Views
2K
Binomial Expansion: Evaluating Coefficient from two binomials
Replies
7
Views
2K
Use of binomial theorem in a sum of binomial coefficients?
Replies
1
Views
2K
Find the ##r^{th}## term of ##(a+2x)^n##
Replies
3
Views
2K
Is the Binomial Theorem Really Worth the Effort to Understand?
Replies
4
Views
2K
Applying binomial theorem to prove combinatorics identity
Replies
8
Views
2K
What are the constants in the binomial series expansion for (1+mx)^-n?
Replies
34
Views
4K
Evaluating Finite Sum: Homework Statement
Replies
5
Views
2K
Coefficient of x^35 in Binomial Theorem Expansion
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top