Question #322895

an=3an-1+(n2+n-2)3n

1
Expert's answer
2022-04-04T16:38:27-0400

an=3an1+(n2+n2)3nHomogeneousequation:an=3an1an=C3nLetan=C(n)3nC(n)3n=3C(n1)3n1+(n2+n2)3nC(n)=C(n1)+n2+n2==C(n2)+(n1)2+(n1)2+n2+n2=...==C(0)+12+22+...+n2+1+2+...+n22...2==C(0)+n(n+1)(2n+1)6+n(n+1)22n==C(0)+n3+3n24n3an=(C(0)+n3+3n24n3)3n=(n3+3n24n+C)3n1a_n=3a_{n-1}+\left( n^2+n-2 \right) 3^n\\Homogeneous\,\,equation:\\a_n=3a_{n-1}\Rightarrow a_n=C\cdot 3^n\\Let\,\,a_n=C\left( n \right) \cdot 3^n\\C\left( n \right) \cdot 3^n=3\cdot C\left( n-1 \right) \cdot 3^{n-1}+\left( n^2+n-2 \right) 3^n\\C\left( n \right) =C\left( n-1 \right) +n^2+n-2=\\=C\left( n-2 \right) +\left( n-1 \right) ^2+\left( n-1 \right) -2+n^2+n-2=...=\\=C\left( 0 \right) +1^2+2^2+...+n^2+1+2+...+n-2-2-...-2=\\=C\left( 0 \right) +\frac{n\left( n+1 \right) \left( 2n+1 \right)}{6}+\frac{n\left( n+1 \right)}{2}-2n=\\=C\left( 0 \right) +\frac{n^3+3n^2-4n}{3}\\a_n=\left( C\left( 0 \right) +\frac{n^3+3n^2-4n}{3} \right) \cdot 3^n=\left( n^3+3n^2-4n+C \right) \cdot 3^{n-1}


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