Answer to Question #178375 in Discrete Mathematics for Dawit workneh

Question #178375
an=-an-1+16an-2+4an-3-48an-4,where a°=0,a1=16,a2=-2,a3=142
1
Expert's answer
2021-04-13T13:58:07-0400

Let us solve "a_n=-a_{n-1}+16a_{n-2}+4a_{n-3}-48a_{n-4}," where "a_0=0,a_1=16,a_2=-2,a_3=142".


Firstly, let us solve the characteristic equation


"k^4=-k^3+16k^2+4k-48"


"k^4+k^3-16k^2-4k+48=0"


"k^4-4k^2+k^3-4k-12k^2+48=0"


"k^2(k^2-4)+k(k^2-4)-12(k^2-4)=0"


"(k^2+k-12)(k^2-4)=0"


"(k-3)(k+4)(k-2)(k+2)=0"


It follows that the characteristoc equation has the following roots:


"k_1=3,\\ k_2=-4,\\ k_3=2,\\ k_4=-2."


Therefore, the general solution is "a_n=c_13^n+c_2(-4)^n+c_32^n+c_4(-2)^n."


Since "a_0=0,a_1=16,a_2=-2,a_3=142", we conclude that


"\\begin{cases}\n0=a_0=c_1+c_2+c_3+c_4\\\\\n16=a_1=3c_1-4c_2+2c_3-2c_4\\\\\n-2=a_2=9c_1+16c_2+4c_3+4c_4\\\\\n142=a_3=27c_1-64c_2+8c_3-8c_4\n\\end{cases}"


"\\begin{cases}\nc_1+c_2+c_3+c_4=0\\\\\n3c_1-4c_2+2c_3-2c_4=16\\\\\n5c_1+12c_2=-2\\\\\n15c_1-48c_2=78\n\\end{cases}"


"\\begin{cases}\n1+c_3+c_4=0\\\\\n10+2c_3-2c_4=16\\\\\nc_2=-1\\\\\nc_1=2\n\\end{cases}"


"\\begin{cases}\nc_4=-2\\\\\nc_3=1\\\\\nc_2=-1\\\\\nc_1=2\n\\end{cases}"


It follows that the final solusion is


"a_n=2\\cdot3^n-(-4)^n+2^n-2\\cdot (-2)^n."



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