Question: What is the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)\cdots (1+x^{100})?$  

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I manually expanded $(1+x)(1+x^2)(1+x^3)...(1+x^{10})$ and calculated the coefficient of $x^9$ as $8$ but i dont know how to solve it without expanding.Please help me.

Answer: 
I think the problem boils down to counting the number of ways to write $9$ as a sum of distinct, positive integers. There is the trivial case of $9=9$, and besides that try by hand to catch all other cases like $1+3+5$, $1+2+6$, etc. I think your best bet is to brute force this problem; one of those times where it's just easier to consider every case than to search for an elegant solution. My reasoning is as follows: 

Think of the product $$\prod\_{n=1}^{100} (1+x^n)$$ as a potential game of telephone where a "message" is passed from one parenthetical quantity to the next via multiplication. In each passage the message can either go to $1$ or $x^n$. For example, we could start at $1$ in $(1+x)$, then go to $1$ in $(1+x^2)$ and $\ldots$ and go to $1$ in $(1+x^{99})$ and lastly go to $x^{100}$ in $(1+x^{100})$ , which is to say we get $1^{99}\cdot x^{100} = x^{100}$ at the end of that particular sequence of multiplication. So we'll expect at least one quantity of $x^{100}$ in the fully expanded product. I'll call "getting $x^{100}$ out of that particular sequence of multiplication" the *result* of that message. More generally, the coefficient $R\_p$ that appears on $x^p$ in the fully expanded product $$\prod\_{n=1}^{100} (1+x^n) = R\_0+R\_1x^1+R\_2x^2+\ldots +R\_px^p +\ldots + R\_{5050}x^{5050}$$ tells us the number of ways our message *resulted* in $x^p$.

Hopefully the analogy is clear. Since we now are interested in the coefficient of $x^9$ here is why I said what I did in my first paragraph. We need to know how many ways there are to pass a message through the product so that it results in $x^9$ by the end. One possible path would be to start at $x$, pass to $x^2$, pass to $1$ in $(1+x^3)$, pass to $1$ in $(1+x^4)$, pass to $1$ in $(1+x^5)$, pass to $x^6$. At this point we *have* to pass to $1$ in the rest of the $(1+x^n)$'s. Otherwise at the end of the message we'll have $x^p$ with $p>9$ at the end. The path I just mentioned results in $$x^1\cdot x^2 \cdot 1^3 \cdot x^6 \cdot 1^{94} = x^9$$ which works because the sum of exponents on each $x$ in the product, $1+2+6$ sums to $9$. Hence you really do just want to count the number of ways to write $9$ as a sum of distinct, positive integers; much easier than computing the entire product. I concur that $R\_9 = 8$.
In every factor of your product, you will need to pick either the $1$ or the $x^i$, so immediatly we see that for all factors with $i > 9$ we need to pick the $1$, otherwise we will get $x^k$ with $k > 9$. So we can just look at the product up till $(1 + x^9)$. Now we look at the number of ways to pick $1$ and $x^i$ here such that we end up with $x^9$. Because all our coefficients are $1$, we do not need to worry about those, and count only the number of ways.

To simplify this, assume you always pick the $1$ if you don't write them down, then all possible combinations are:
$x^9, x^1\cdot x^8, x^2\cdot x^7, x^3\cdot x^6, x^4\cdot x^5, x^1\cdot x^2\cdot x^6, x^1\cdot x^3\cdot x^5, x^2\cdot x^3\cdot x^4$ which gives us $8$ as the final result.

It is not hard to see that this is equivalent to summing numbers $1, \ldots 9$ to $9$ using every number only once, and then counting the number of summations.