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If $f(x) = (x-1)\cdot (x-2)\cdot (x-3)\cdot (x-4)\cdot ........(x-100)\;,$ Then Coefficient of $x^{99}$ and

Coefficient of $x^{98}$ and Coefficient of $x^{97}$ in $f(x).$

$\bf{My\; try::}$ We can write $f(x)$ as

$$\displaystyle f(x) = x^{100}-\left(\sum_{i=1}^{100}i\right)x^{99}+\left(\mathop{\sum^{100}\sum^{100}}_{i=1\ \ j=1\ i<j}i\cdot j\right)x^{98}-\left(\mathop{\sum^{100}\sum^{100}\sum^{100}}_{i=1\ j=1\ k=1\ i<j<k}i\cdot j\cdot k\right)x^{97}+..$$

So $\bf{Coefficients}$ of $$\displaystyle x^{99} = -\sum_{i=1}^{100}\left(1+2+3+......+100\right) = -\frac{100\cdot 101}{2} = -5050$$

Similarly Coeff. of $$\displaystyle x^{98} = \mathop{\sum^{100}\sum^{100}}_{i=1\ \ j=1\ i<j}i\cdot j = \frac{1}{2}\left[\left(\sum_{i=1}^{100}i\right)^2-\sum_{i=1}^{100}i^2\right] = \frac{1}{2}\left[\left(\frac{100\cdot 101}{2}\right)^2-\frac{100\cdot 101\cdot 201}{6}\right]$$

But I did not understand How can i calculate Coeff. of $\displaystyle x^{97} =-\mathop{\sum^{100}\sum^{100}\sum^{100}}_{i=1\ j=1\ k=1\ i<j<k}i\cdot j\cdot k$

Help me, Thanks

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juantheron
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1 Answers1

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For the coefficient of $x^{97}$ you can apply a method similar to the one you used to find the coefficient of $x^{98}$. You already noted that the coefficient $C$ of $x^{97}$ is $$C:=-\mathop{\sum^{100}\sum^{100}\sum^{100}}_{i=1\ j=1\ k=1\ i<j<k}i\cdot j\cdot k=-\sum_{i=1}^{98}\sum_{j=i+1}^{99}\sum_{k=j+1}^{100}i\cdot j\cdot k.$$ We can compute $C$ from the identity $$\left(\sum_{i=1}^{100}i\right)^3=-2\sum_{i=1}^{100}i^3+3\left(\sum_{i=1}^{100}i^2\right)\left(\sum_{j=1}^{100}j\right)-6C.$$ Rewriting the above yields \begin{eqnarray*} C&=&-\frac{1}{6}\left[\left(\sum_{i=1}^{100}i\right)^3+2\sum_{i=1}^{100}i^3-3\left(\sum_{i=1}^{100}i^2\right)\left(\sum_{j=1}^{100}j\right)\right],\\ &=&-\frac{1}{6}\left[\left(\tfrac{1}{2}\cdot100\cdot101\right)^3+2\cdot\left(\tfrac{1}{4}\cdot100^2\cdot101^2\right)-3\cdot\left(\tfrac{1}{6}\cdot100\cdot101\cdot201\right)\cdot\left(\tfrac{1}{2}\cdot100\cdot101\right)\right]\\ &=&-20618771250, \end{eqnarray*} modulo miscalculations.

Servaes
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  • According to Wolfram Alpha, $C=-20618771250$. – mathlove Aug 16 '15 at 13:10
  • I found $\frac{1}{48}\left(n-2\right)\left(n-1\right)n^{2}\left(n+1\right)^{2}$ and substitution $n=100$ says that wolfram is correct. Though, the expression for $C$ looks okay to me. – drhab Aug 16 '15 at 13:13
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    @Servaes : You treat $C$ as $-C$ (note that $C$ is negative). Your answer seem correct except the very last calculation. – mathlove Aug 16 '15 at 13:27
  • Is there a general method available to find the coefficient of any term like suppose,find the coefficient of $x^{81}$.@Servaes – Brahmagupta Feb 06 '16 at 08:10