xét $(1+x)(1+2x)...(1+2013x)=a_{0}+a_{1}x+a_{2}x^{2}+....+a_{2013}x^{2013}$
Tính $a_{2}+\frac{1}{2}(1^{2}+2^{2}+...+2013^{2}$
ban oi hinh nhu ban nham de roitinh a2 1/2*(1^2*2^2*3^3.......) trong ngoac la dau nhan moi dung –  thien_su_skygardent 06-12-15 11:58 AM
sao ma kho ghi vao qua –  thien_su_skygardent 06-12-15 11:27 AM
a/ la j z –  thien_su_skygardent 06-12-15 10:42 AM
Xét $F(x)=\prod_{n=1}^{2013}(1+nx)$
$F'(x)=\sum_{i=1}^{2013}[\frac{i}{1+ix}.\prod_{n=1}^{2013}(1+nx)] $
$F''(x)=[2(1+3x)(1+4x)...(1+2013x)+3(1+2x)(1+4x)...(1+2013x)+...+2013(1+2x)(1+3x)...(1+2012x)]+[2(1+2x)(1+4x)...(1+2013x)+2.3(1+x)(1+4x)...(1+2013x)+...+2.2013(1+x)(1+3x)...(1+2012x)]+...+[2013(1+2x)(1+3x)...(1+2012x)+2013.2(1+x)(1+3x)...(1+2012x)+...+2013.2012(1+x)(1+2x)...(1+2011x)]$
$\Rightarrow F'(0)=1(2+3+4+...+2013)+2(1+3+4+...+2013)+...+2013(1+2+3+4+...+2012)$
$=(1+2+3+...+2013)^2-(1^2+2^2+3^2+...+2013^2)$ (1)
Mặt khác: $F(x)=\sum_{n=0}^{2013}a_{n}x^n $
$F'(x)=\sum_{n=1}^{2013}na_{n}x^{n-1} $
$F''(x)=2a_2+2.3a_3x+...+2013.2012a_{2013}x^{2011}$
$\Rightarrow F'(0)=2a_2$ (2)
Từ (1) và (2), suy ra: $a_2=\frac{1}{2}(1+2+3+...+2013)^2-\frac{1}{2}(1^2+2^2+3^2+...+2013^2)$
Như vậy: $a_2+\frac{1}{2}(1^2+2^2+3^2+...+2013^2)=\frac{1}{2}(1+2+3+...+2013)^2=\frac{1}{2}(\frac{2013.2014}{2})^2$
ôi. hoa mắt thật –  rang 09-12-15 09:21 PM
cho x=0 ->Ao=0=2011C0 ->A2=2011C2
a2+12(12+22+...+20132
=2011C2 +$\frac{1}{2}(1^{2}+2^{2}+....+2013^{2})$



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