Question on binomial coefficient

Question

What will be the coefficient of $x^{99}$ in the following: $(x+1)(x+2)(x+3).....(x+100)$ Please help. I am stuck on this.

score 1 · Answer 1 · answered Nov 27 '17 at 20:07

1

You will need to pick $x$ from every bracket apart from one giving \begin{eqnarray*} (x+1)(x+2)(x+3)\cdots(x+100)=x^{100}+ x^{99}\color{red}{\left(\sum_{i=1}^{100}i\right)}+\cdots \end{eqnarray*}

answered Nov 27 '17 at 20:07

Donald Splutterwit

36,613
2
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this is the same as choosing combinations of (100-1) factors from 100 – Abr001am Nov 27 '17 at 20:08

score 0 · Accepted Answer · answered Nov 27 '17 at 20:27

0

Zeroes of the given polynomial are $-1,-2,...-100$. By Viete formulas for $$ax^{100} + bx^{99}+....=0$$ we have $$ x_1+x_2+...x_{100} = -{b\over a}$$ so $$-b = -1-2-...-100 = -{100\cdot 101\over 2} \Longrightarrow b= 5050$$

answered Nov 27 '17 at 20:27

nonuser

90,026

You bastard, you did it again! – nonuser Nov 28 '17 at 20:22

Question on binomial coefficient

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