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

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.

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......

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 \((\frac{n!}{0!.n!.0!....0!})\) \(1^{0}.x^{n}.x^{2.0}. ... .x^{r.0}\)

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.... :)

## Comments

Sort by:

TopNewestthe 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 itLog in to reply

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.

Log in to reply

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......Log in to reply

Log in to reply

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 tonand all otherk'si.e\(k_{1}, k_{3}, ..., k_{r+1}\)should beZero.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

\(\boxed{1}\)Log in to reply

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

Thank you

Log in to reply

But sum of all

k'sshould also be equal ton.but in the case you have thought How the sum of all

k'swill ben.Log in to reply

\(k_{1} = 1\)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

Log in to reply