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Finding product of terms taken $3$ at a time in $\displaystyle \prod^{100}_{r=1}(x+r)$

Try:

$$\displaystyle \prod^{100}_{r=1}(x+r)=x^{100}+(1+2+3+\cdots +100)x^{99}+(1\cdot 2+1\cdot 3+\cdots+100\cdot 99)x^{98}+(1\cdot 2\cdot 3+2\cdot 3 \cdot 4+\cdot\cdots+98\cdot 99\cdot 100)x^{98}+\cdots$$

for $1$ at a time (Coefficient of $x^{99}$) is $\displaystyle \sum^{100}_{i=1}i = 50\cdot 101$

for $2$ at a time (Coefficient of $x^{98}$) is $\displaystyle \sum^{100}_{i=1}\sum^{100}_{j=1\;, (1\leq i<j \leq 100)}i \cdot j= \frac{1}{2}\bigg[\bigg(\sum^{100}_{i=1}i\bigg)^2-\sum^{100}_{i=1}i^2\bigg]$

But could some help me how to find coefficient of $x^{97},$ thanks

DXT
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3 Answers3

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The coefficient of $x^{97}$ is $$ \begin{align} \sum_{i=1}^{100}\sum_{j=1}^{i-1}\sum_{k=1}^{j-1}ijk &=\sum_{i=1}^{100}\sum_{j=1}^{i-1}\sum_{k=1}^{j-1}ij\binom{k}{1}\\ &=\sum_{i=1}^{100}\sum_{j=1}^{i-1}i\,((j-2)+2)\binom{j}{2}\\ &=\sum_{i=1}^{100}\sum_{j=1}^{i-1}i\left[3\binom{j}{3}+2\binom{j}{2}\right]\\ &=\sum_{i=1}^{100}i\left[3\binom{i}{4}+2\binom{i}{3}\right]\\ &=\sum_{i=1}^{100}\left[3((i-4)+4)\binom{i}{4}+2((i-3)+3)\binom{i}{3}\right]\\ &=\sum_{i=1}^{100}\left[15\binom{i}{5}+12\binom{i}{4}+8\binom{i}{4}+6\binom{i}{3}\right]\\ &=15\binom{101}{6}+20\binom{101}{5}+6\binom{101}{4}\\[9pt] &=20618771250 \end{align} $$

robjohn
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With some manipulation (using inclusion exclusion principle) you can write the sum as:

$$\left(\sum_{i=1}^{100} i\right)^3 = \binom{3}{1}\left(\sum_{i=1}^{100} i^2\right)\left(\sum_{i=1}^{100} i\right) - 2\left(\sum_{i=1}^{100}i^3\right) + \binom{3}{1} \binom{2}{1}S $$

Where $S$ is the required sum.

jonsno
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  • how can we use inclusion exclusion please explain me, thanks – DXT Dec 28 '17 at 05:30
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    @Durgesh Thats what you did in $a_{97}$ right? In $\left(\sum_{i=1}^{100}i\right)^3$ There are different series generated. But our required series is embedded there. We need to subtract some series to get to our required series. For more info see PIE – jonsno Dec 28 '17 at 05:34
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If you try to expand the multiplications, you'll find that to produce terms with $x^{98}$, you'll need $2$ distinct constant factors (from the set $\{1,2,\dots,100\})$ and $98$ number of $x$ factors. So the sum of these terms, $a_{98}$, would be $$a_{98}=\sum_{n=1}^{100}\sum_{m=n+1}^{100}mn$$

How can we find this sum? Note that $$(1+2+…100)^2= 1^2+2^2+… 100^2 + 2a_{98}$$ $$\implies a_{98} = \frac12 \left[(5050)^2- \sum_{i=1}^{100} i^2 \right]$$ $$\implies \boxed{ a_{98} = 12582075}$$

  • thanks rohan i mean coefficient of $x^{97}$ – DXT Dec 28 '17 at 05:29
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    @DurgeshTiwari See if this MSE Question helps: https://math.stackexchange.com/questions/1399182/coeff-of-x97-in-fx-x-1-cdot-x-2-cdot-x-3-cdot-x-4-cdot?rq=1 –  Dec 28 '17 at 05:33
  • @Rohan that makes this question duplicate imo, i got same answer as in that post – jonsno Dec 28 '17 at 05:37
  • @DurgeshTiwari Well fine if it is a duplicate, then cheers for your effort! –  Dec 28 '17 at 05:38