Cho khai triển P(x)=(1+x)(1+2x)...(1+2017x)=a0+a1x+...+a2017x^2017

* Hướng dẫn giải

Đáp án D

Ta có $1^2+2^2+3^2+\mathrm{...}+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

và $1+2+3+\mathrm{...}+n^2=\frac{n\left(n+1\right)}{2}$

Xét $\left(1+x\right)\left(1+2x\right)\mathrm{...}\left(1+nx\right)\Rightarrow$ Hệ số của $x^2$ là

 $a_2=1.\left(2+3+\mathrm{...}+n\right)+2.\left(3+4+\mathrm{...}+n\right)+\mathrm{...}+\left(n-1\right)n$

$=1.\left[\left(1+2+\mathrm{...}+n\right)-1\right]+2.\left[\left(1+2+\mathrm{...}+n\right)-\left(1+2\right)\right]+\mathrm{...}+\left(n-1\right).\left[\left(1+2+\mathrm{...}+n\right)-\left(1+2+\mathrm{...}+n-1\right)\right]$

$=\sum _{k=1}^{n}k\times \left[\frac{n\left(n+1\right)}{2}-\frac{k\left(k+1\right)}{2}\right]$

$=\frac{1}{2}\sum _{k=1}^{n}k\times \left[\left(n^2+n\right)-\left(k^2+k\right)\right]$

$=\frac{1}{2}\sum _{k=1}^n\left[\left(n^2+n\right)k-\left(k^3+k^2\right)\right]$

$=\frac{1}{2}=\frac{{\left(n^2+n\right)}^2}{8}-\frac{n\left(n+1\right)\left(2n+1\right)}{12}$

$\left[\frac{{\left(n^2+n\right)}^2}{2}-\frac{{\left(n^2+n\right)}^2}{4}-\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]$

Vậy $T=\frac{{\left(n^2+n\right)}^2}{8}$

$\stackrel{n-2017}{\to }T=\frac{{\left(2017.2018\right)}^2}{8}=\frac{1}{2}{\left(\frac{2017.2018}{2}\right)}^2$