Question

这题咋做？要详解

Answer 1

错位相减法：an=(2n-1)×3^n
Sn=1×3^1+3×3^2+5×3^3+…+(2n-3)×3^(n-1)+(2n-1)×3^n ①
那么3Sn=1×3^2+3×3^3+…+(2n-3)×3^n+(2n-1)×3^(n+1) ②
①-②：-2Sn=1×3^1+2×3^2+2×3^3+…+2×3^n-(2n-1)×3^(n+1)
=3+2×(3^2+3^3+…+3^n)-(2n-1)×3^(n+1)
=3+2×9×[1-3^(n-1)]/(1-3)-(2n-1)×3^(n+1)
=-6-2(n-1)×3^(n+1)
∴Sn=3+(n-1)×3^(n+1)

Answer 2

Sn=1·3+3·3²+5·3³+...+(2n-1)·3ⁿ
3Sn=1·3²+3·3³+...+(2n-3)·3ⁿ+(2n-1)·3ⁿ⁺¹
Sn-3Sn=-2Sn=3+2·3²+2·3³+...+2·3ⁿ-(2n-1)·3ⁿ⁺¹
=2·(3+3²+...+3ⁿ) -(2n-1)·3ⁿ⁺¹-3
=2·3·(3ⁿ-1)/(3-1) -(2n-1)·3ⁿ⁺¹-3
=(2-2n)·3ⁿ⁺¹-6
Sn=(n-1)·3ⁿ⁺¹+3