Binomial theorem problem

[BrianDaMan]

The question asks me to prove that the second last term in the expansion of (a+b)^n is n/5(n-5) x b^n. The answer says like the photo but i don’t understand that they got (n under 1)a^1b^(n-1). Isn’t it (n under n - 1) if it’s the second last term? Am i getting wrong? Or is the answer wrong?

thanks

 

[Levido]

Here’s what might help you

The Binomial expansion for [MATH]nCr[/MATH] in this situation, (nC(n-1)) is [MATH]\frac{n!}{r!(n-r)!}[/MATH]. This is, in your expression, n under 1 which I’ve never seen before so I won‘t use.

A key property of this is that it is symmetrical about its middle term, and if there is an even number of terms, the two central terms are equal. Ill illustrate this below

Let’s r=1: [MATH]nCr=\frac{n!}{1*(n-1!)}[/MATH]
Now let’s say r=n-1: [MATH]nCr=\frac{n!}{(n-(n-1)!}=\frac{n!}{(n-1)!*1!}[/MATH]
As demonstrated, [MATH]nC1=nC(n-1)[/MATH]
I’ll look through the rest of the issues and answer below

 

[Levido]

Please provide the full question as it looks like there is some relation between a and b

 

[JeffM]

It is tough to say whether an answer is correct when you do not give us the complete problem.

[MATH](a + b)^n = \sum_{j=0}^n \dbinom{n}{j} * a^{(n-j)} * b^j.[/MATH]
What does j equal on the next to last term. Clearly, n - 1

[MATH]t_{n-1} = \dbinom{n}{j} * a^{(n-j)} * b^j = \dbinom{n}{n-1} * a^{\{n - (n - 1)\}} * b^{(n-1)} \implies[/MATH]
[MATH]t_{n-1} = \dfrac{n!}{(n - 1)! * \{n - (n - 1)\}!} * a^1 * b^{(n-1)} = \dfrac{n * (n - 1)!}{(n - 1)! * 1!} * ab^{(n-1)} \implies[/MATH]
[MATH]t_{n-1} = nab^{(n-1)}.[/MATH]
As for the rest, it could be right or wrong. How can we know when you do not provide the exact and complete statement of the problem AS WE REQUEST at

https://www.freemathhelp.com/forum/threads/guidelines-summary.112086/#post-433156

 

[Steven G]

_nC_x=_nC_n-x.

So _nC_n-1 = _nC₁

 

[lex]

There'll be some relationship between a couple of terms e.g. [MATH]t_5=30t_6[/MATH]. But I guess we'll never know.

 

[HallsofIvy]

You can't prove that- it is not true! Every term in the expansion of \(\displaystyle (a+ b)^n\) is a constant times \(\displaystyle a^ib^j\) with i+ j= 1.

One reason what you have is not true is that the problem is about \(\displaystyle (a+ b)^n\) but you have "x"! That is, I suppose, a simple typo. More importantly the exponents are 1 and n and they add to n+1, not n.

 

[lex]

You can't prove that- it is not true! Every term in the expansion of \(\displaystyle (a+ b)^n\) is a constant times \(\displaystyle a^ib^j\) with i+ j= 1.

One reason what you have is not true is that the problem is about \(\displaystyle (a+ b)^n\) but you have "x"! That is, I suppose, a simple typo. More importantly the exponents are 1 and n and they add to n+1, not n.

Perhaps you would consider a couple of the points you are making?

i+j=1?

I think the 'x' you refer to may be meant to represent multiplication.
Possibly that would take care of the exponents adding to n+1?