高中数学数列：已知an=(2n+1)(-3)^n，求前n项和sn
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Sn=3·(-3)+5·(-3)²+7·(-3)³+...+(2n+1)·(-3)ⁿ
-3Sn=3·(-3)²+5·(-3)³+...+(2n-1)·(-3)ⁿ+(2n+1)·(-3)ⁿ⁺¹
Sn-(-3Sn)=4Sn=3·(-3)+2·(-3)²+2·(-3)³+...+2·(-3)ⁿ-(2n+1)·(-3)ⁿ⁺¹
=2·[(-3)+(-3)²+...+(-3)ⁿ]-(2n+1)·(-3)ⁿ⁺¹-3
=2·(-3)·[(-3)ⁿ-1]/(-3-1) -(2n+1)·(-3)ⁿ⁺¹-3
=-[(4n+3)·(-3)ⁿ⁺¹+9]/2
Sn=-[(4n+3)·(-3)ⁿ⁺¹+9]/8
-3Sn=3·(-3)²+5·(-3)³+...+(2n-1)·(-3)ⁿ+(2n+1)·(-3)ⁿ⁺¹
Sn-(-3Sn)=4Sn=3·(-3)+2·(-3)²+2·(-3)³+...+2·(-3)ⁿ-(2n+1)·(-3)ⁿ⁺¹
=2·[(-3)+(-3)²+...+(-3)ⁿ]-(2n+1)·(-3)ⁿ⁺¹-3
=2·(-3)·[(-3)ⁿ-1]/(-3-1) -(2n+1)·(-3)ⁿ⁺¹-3
=-[(4n+3)·(-3)ⁿ⁺¹+9]/2
Sn=-[(4n+3)·(-3)ⁿ⁺¹+9]/8
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