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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.

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

2 Answers 2

6
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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.

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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 $1$).

The sequences we care about

  • 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.

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