What is the co-efficient of \(x^n\) of the following expression:

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

What is the co-efficient of \(x^n\) of the following expression:

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

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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 it – Vaibhav Reddy · 3 years, 9 months agoLog in to reply

This is much trickier when \( r < < n\), and you'd likely need to use the Principle of Inclusion and Exclusion. – Calvin Lin Staff · 3 years, 9 months ago

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\(r<<n\),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...... – Vaibhav Reddy · 3 years, 9 months agoLog in to reply

– Vaibhav Reddy · 3 years, 9 months ago

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

\(\boxed{1}\)– Sandeep Saurav · 3 years, 9 months agoLog in to reply

\(Multinomial\)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 – Vaibhav Reddy · 3 years, 9 months ago

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

k'swill ben. – Sandeep Saurav · 3 years, 9 months agoLog 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 – Vaibhav Reddy · 3 years, 9 months ago

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