# Find the coefficient of ${x^9}$ in the expansion of $(1 + x)( 1 + x^2)( 1 + x^3)..(1 + x^{100})$ [duplicate]

Find the coefficient of $${x^9}$$ in the expansion of $$\left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^3}} \right)..\left( {1 + {x^{100}}} \right)$$. The official answer is 8.

How do I find the general term,

Dividing the above equation by $$(1-x)$$ is not generating the required result.

• Expanding the product will result in $2^{100}$ terms of the form $x^k$. You want to count how many of them are $x^9$. Dec 25, 2020 at 15:45
• isnt this an IIt JEE question if i am not mistaken? Dec 25, 2020 at 15:52
• Yes this is IIT JEE Question Dec 25, 2020 at 18:01
• I'm really not sure but I doubt if we could use partition of integer
– user960916
Oct 4, 2021 at 6:01

Hint:

\begin{align}9=9+0\\=8+1\\=7+2\\=6+3\\=6+2+1\\=5+4\\=5+3+1\\=4+3+2\end{align}

We don't have to worry about 4 summands, since $$1+2+3+4>9$$. There is no known closed form for the general term.

An "x^9" occurs in this expression exactly when a sequence of powers of $$x$$s add up to a total exponent of $$9$$. For instance, the first and eighth terms contribute $$x$$ and $$x^8$$, for a total of $$x^9$$ (where in all other terms you multiply by $$1$$).

• start at 1
• are increasing
• Have no two exponents the same

So now you can just start writing them down:

9
1,8
2,7
3,6
4,5
1,2,6
1,3,5
2,3,4


... and there are eight of those.