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Anyone please teach me how to solve such questions... I am thinking of solving it using binomial theory but still no idea how to do it..

Plz help

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    $\begingroup$ Hint: the polynomial has roots $1, 2, \cdots 100$. By Vieta's formulas the coefficient of, for example, $x^{99}$ is $-(1+2+\cdots 100)$. Try to use the next formula for the coefficient of $x^{98}$. $\endgroup$ – dxiv Nov 12 '16 at 6:49
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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 $$a_{98}=\sum_{n=1}^{100}\sum_{m=n+1}^{100}mn$$ which you can rewrite as \begin{aligned} a_{98}&=\frac12\left(\sum_{n=1}^{100}\sum_{m=1}^{100}mn - \sum_{n=1}^{100}n^2\right)\\ &=\frac12\left(\sum_{n=1}^{100}n\sum_{m=1}^{100}m - \sum_{n=1}^{100}n^2\right)\\ &= \frac12\left(5050^2-338350\right)\\ &=12582075 \end{aligned}

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$$(x-1)(x-2)(x-3)…(x-100)=x^{100} (1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{100}{x})$$

Denote $y=\frac{1}{x}$. Then $$x^{100} (1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{100}{x}) =x^{100} (1-y)(1-2y)...(1-100y)$$ and the problem becomes: Find the coefficient of $y^2$ in $$P(y)=(1-y)(1-2y)...(1-100y)$$ This coefficient is $$\frac{P''(0)}{2!}$$

Derivating twice and plugging $x=0$ we get $$P''(0)=\sum_{k=1}^{100}\sum_{j=1, j\neq k}^{100} kj$$ thus your coefficient is $$\frac{1}{2} \sum_{k,j =1 , k\neq j}^{100} kj= \frac{1}{2} \left( \left( \sum_{k=1}^{100} k \right)^2 - \left( \sum_{k=1}^{100} k^2 \right) \right)$$

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More generally, for any problem of this kind, you can use the Stirling numbers of the first kind $S_1(n,k)$. One has $$ x(x-1)\cdots (x-n+1)=\sum_{k=0}^n\,(-1)^{n-k}S_1(n,k)x^k\ . $$ Here, the desired coefficient is then $(-1)^2S_1(100,98)=S_1(100,98)$ which can be computed through its mixed generating function.

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