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The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$.

I tried the following concept, how to sum 9 using 1-9 only without repetition.

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

6:1+2+6 ,7:1+3+5, 8:2+3+4

The answer is 8.

How will it be solved using Binomial theorem.

Samar Imam Zaidi
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    I wouldn't solve it with binomial theorem, I would solve it your way. Since we're not dealing with a factor raised to a power, binomial theorem is a little useless here. – Theo Bendit Sep 19 '18 at 14:09
  • Binomial theorem concerns expansions of expressions of the form $(a+b)^n$. Here we have $(a+b_1)(a+b_2)...$, for which binomial theorem is not usable since none of the $b_i$ are equal. Instead, a matching of the powers like you have done is the best way of doing things. – Sarvesh Ravichandran Iyer Sep 19 '18 at 14:10
  • Note that $$(1+x)(1+x^2)\cdots (1+x^{100}) =\ (1+x)(1+x^2)\cdots (1+x^9) \quad + \quad x^{10}\times P(x),$$ where $P(x)$ is polynomial of $x$;

    so we can focus on $9$ first multipliers.

     – Oleg567 Sep 19 '18 at 14:20

2 Answers2

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Yes we can count directly the cases that is

  • $x^9$
  • $x\cdot x^8$
  • $x\cdot x^2\cdot x^6$
  • $x\cdot x^3\cdot x^5$
  • $x^2\cdot x^7$
  • $x^2\cdot x^3\cdot x^4$
  • $x^3\cdot x^6$
  • $x^4\cdot x^5$

to obtain $8$.

user
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As Theo Bendit has already commented this is not solved by Newton's binomial theorem. The solution is the same way that is used to prove Newton's bionomian theorem that is the proper use of combinatorial principles. Let $$ p(x)=(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)(1+x^6)(1+x^7)(1+x^8)(1+x^9) $$ We have \begin{align} p(x) =& 1 +\sum_{1\leq i_1\leq 9}x^{i_1} +\sum_{1\leq i_1<i_2\leq 9}x^{i_1}\cdot x^{i_2} +\sum_{1\leq i_1<i_2<i_3\leq 9}x^{i_1}\cdot x^{i_2}\cdot x^{i_3} + \\ \\ &\qquad +\sum_{1\leq i_1<i_2<i_3<i_4\leq 9}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}\cdot x^{i_4} +\ldots +\sum_{1\leq i_1<\cdots <i_9\leq 9}x^{i_1}\cdot \dots \cdot x^{i_9} \end{align} or for $m_1=\max\{i_1: 1\leq i_1\leq 9\}$, $m_2=\max\{i_1+i_2: : 1\leq i_1<i_2\leq 9\}$,$\ldots$, $m_9=\max\{i_1+\ldots+i_9: 1\leq i_1<\ldots <i_9\leq 9\}$ \begin{align} p(x) =& 1 +\sum_{k=1}^{m_1}\sum_{i_1= k}x^{i_1} +\sum_{k=1+2}^{m_2}\sum_{\substack{ i_1<i_2 \\i_1+i_2=k}}x^{i_1}\cdot x^{i_2} +\sum_{k=1+2+3}^{m_3}\sum_{\substack{ i_1<i_2<i_3 \\i_1+i_2+i_3=k}}x^{i_1}\cdot x^{i_2}\cdot x^{i_3} +\ldots \\ \\ &\qquad +\sum_{k=1+2+3+4}^{m_4}\;\;\sum_{i_1<i_2<i_3<i_4}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}\cdot x^{i_4} + \qquad\ldots+\sum_{k=1+\ldots+9}^{m_9}\sum_{\substack{ i_1<\cdots <i_9\\ i_1+\ldots+i_9=k}}x^{i_1}\cdot \dots \cdot x^{i_9} \end{align} All possible powers of $ x $ equal to $ x ^ 9 $ are generated by the first three summations. Let $ c_9 $ be the coefficient of $ x ^ 9 $ in the polynomial $ p (x) $. So \begin{align} c_9\cdot x^9 =& \sum_{\substack{1\leq i_1\leq 9\\ i_1=9}}x^{i_1} +\sum_{\substack{1\leq i_1<i_2\leq 9\\ i_1+i_2=9}}x^{i_1}\cdot x^{i_2} +\sum_{\substack{1\leq i_1<i_2<i_3\leq 9\\ i_1+i_2+i_3=9}}x^{i_1}\cdot x^{i_2}\cdot x^{i_3} \\ =& 1\cdot x^9+1\cdot x^1\cdot x^8+1\cdot x^2\cdot x^7+1\cdot x^3\cdot x^6+1\cdot x^4\cdot x^5 \\ &+1\cdot x^1\cdot x^2\cdot x^6+1\cdot x^1\cdot x^3\cdot x^5+1\cdot x^2\cdot x^3\cdot x^4 \\ &=8x^9 \end{align}

Elias Costa
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