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# Help me to find co-efficient of $$x^n$$

What is the co-efficient of $$x^n$$ of the following expression:

$$(1+x+x^2+x^3+ \dots + x^r)^n$$

Note by Fazla Rabbi
4 years, 2 months ago

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the function inside the parenthesis is 1+$$x$$+$$x^{2}$$ + $$x^{3}$$ + ........ + $$x^{r}$$ can be converted into $$\frac{x^{r+1}-1}{(x-1)}$$ ... (well that's a formula) .....I guess you can do the remaining.....I'd like to keep it short..If you don't know the formula for $$(x-1)^{-n}$$....you can google it

- 4 years, 2 months ago

You need to be careful with your statements, and flesh out all the details. You cannot just switch it over to $$\frac{1}{1-x}$$, which seems to be what you want to do. Note that your answer must depend on $$r$$ and $$n$$.

This is much trickier when $$r < < n$$, and you'd likely need to use the Principle of Inclusion and Exclusion.

Staff - 4 years, 2 months ago

Thank you sir.....but sorry that i couldn't understand what you meant....... in case of $$r<<n$$,I don't find anything wrong in using that formula....

Sorry....I shouldn't have neglected the question...thanks again sir and pardon me

Neglecting the negative sign for a while.... $$1+nx+ \frac{(n)(n-1)}{2}x^{2} + ..........=(1-x)^{-n}$$

$$(x^{r+1} - 1)^{n} = x^{(r+1)(n)} - nx^{(r+1)(n-1)} + ... - + - + - ...... + (-1)^{n}$$

Now we can calculate the value of the coefficient if we know the value of $$r$$...

It seems I can't find the value of the co-efficient of $$x^{n}$$ without assuming the value of $$r$$ ......I cannot find the value in terms of $$r$$.....I guess somebody is going to clear that part for me......

- 4 years, 2 months ago

now it's kinda happy to type using the formatting guide.....earlier it was harder for me to express the questions

- 4 years, 2 months ago

Using Multinomial Theorem :

i.e. for $$(x_{1} + x_{2} + x_{3} + … + x_{n})^{m}$$

the coefficient of any term can be computed by the formula :

$$(\frac{m!}{k_{1}!.k_{2}!.k_{3}!....k_{n}!} )$$$$x_{1}^{k_{1}}.x_{2}^{k_{2}}.x_{3}^{k_{3}}. ... .x_{n}^{k_{n}}$$ and
$$k_{1}+k_{2}+k_{3}+...+k_{n}$$=m.

so, the coefficient of $$x^{n}$$ in the expression $$(1+x+x^{2}+...+x^{r})^{n}$$ can be calculated as follows :-

The coefficient of general expression will be $$(\frac{n!}{k_{1}!.k_{2}!.k_{3}!....k_{r+1}!} )$$.$$1^{k_{1}}.x^{k_{2}}.x^{2.k_{3}}. ... .x^{r.k_{r+1}}$$ and for $$x^{n} k_{2}$$ should be equal ton and all other k's i.e $$k_{1}, k_{3}, ..., k_{r+1}$$ should be Zero.

so, the coefficient will be $$(\frac{n!}{0!.n!.0!....0!})$$ $$1^{0}.x^{n}.x^{2.0}. ... .x^{r.0}$$

so, the answer will be $$\boxed{1}$$

- 4 years, 2 months ago

That was brilliant of you to use the $$Multinomial$$ theorem....but unfortunately the answer can't be found by taking all the coefficients except one equal to 0......

It goes longer than you expected......

What if I think* $$k_{3} = 1$$ and $$k_{2} = n-2$$ *.....see that's also a possibility

There are many more possibilities.....You should check them as well.... :)

Thank you

- 4 years, 2 months ago

But sum of all k's should also be equal to n.

but in the case you have thought How the sum of all k's will be n.

- 4 years, 2 months ago

oh...in that case I would just take $$k_{1} = 1$$ and all the other equal to $$0$$....

So we get $$k_{1}+k_{2}+k_{3}=n$$...

I guess you got what i was talking about......that's what matters

- 4 years, 2 months ago