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Here's my question : "I have to find the coefficient of $x^8$ in the expansion of : $$1 + (1 + x) + (1 + x)^2 + ..... + (1 + x)^n ;n\geq 8$$Now, I don't know how to apply the binomial theorem to a sum of multiple binomial expansions. Can anyone at least give some hints....??"

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Another way that you can do this is to realize that the given sum is geometric: recall that $$ \sum_{k=0}^{n}a\cdot r^k=\frac{a(1-r^{n+1})}{1-r}. $$ So, taking $a=1$ and $r=1+x$, we find that $$ \sum_{k=0}^{n}(1+x)^k=\frac{1-(1+x)^{n+1}}{1-(1+x)}=\frac{(1+x)^{n+1}-1}{x}. $$ Now, you know that $$ (1+x)^{n+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}x^k, $$ and therefore $$ (1+x)^{n+1}-1=\sum_{k=1}^{n+1}\binom{n+1}{k}x^k; $$ so, as a result, $$ \frac{(1+x)^{n+1}-1}{x}=\sum_{k=1}^{n+1}\binom{n+1}{k}x^{k-1}=\sum_{j=0}^{n}\binom{n+1}{j+1}x^j, $$ by reindexing with $j:=k-1$. Thus, the coefficient by $x^8$ (assuming that $n\geq 8$ so that there IS such a coefficient) is simply $\binom{n+1}{9}$.

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You can just calculate the coefficient of $x^8$ in each $(1+x)^i$ and then sum from $i=0$ to $i=n$. The coefficient of $x^8$ in $(1+x)^i$ is $\binom i8$. Summing from $0$ to $n$ yields: $$ \sum_{i=0}^n\binom i8=\sum_{i=8}^n\binom i8 $$ We have the general identity $$ \sum_{i=0}^n\binom ik=\binom{n+1}{k+1} $$ Therefore, we can write our result (where $k=8$) as $$ \binom {n+1}9 $$

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    $\begingroup$ And the final result is $\left( \begin{array}{c} n+1 \\ 9 \end{array} \right)$. $\endgroup$ – Przemo Feb 7 '14 at 13:27
  • $\begingroup$ @Przemo (+1): you might find \binom{n+1}{9} or {n+1 \choose 9} easier to type in LaTeX, rather than an array $\endgroup$ – Henry Feb 7 '14 at 13:50

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