Answer to Question #117123 in Combinatorics | Number Theory for Priya

Question

Show that the sequence 〖{a〗_n} is a solution of the recurrence relation a_n=a_(n-1)+〖2a〗_(n-2)+2n-9 if
(i)〖 a〗_n=-n+2
(ii) a_n=〖5(-1)〗^n-n+2
(iii) 〖3(-1)〗^n+2^n-n+2
(iv)〖7.2〗^n-n+2

Answer 1

(i) Given $a_n=-n+2.$ Prove

$$a_n=a_{n-1}+2a_{n-2}+2n-9,n\geq2$$ $$a_{n-1}=-(n-1)+2=-n+3$$ $$a_{n-2}=-(n-2)+2=-n+4$$

$$a_{n-1}+2a_{n-2}+2n-9=-n+3-2n+8+2n-9=$$ $$=-n+2=a_n,n\geq2$$

(ii) Given $a_n=5(-1)^n-n+2.$ Prove

$$a_n=a_{n-1}+2a_{n-2}+2n-9,n\geq2$$ $$a_{n-1}=5(-1)^{n-1}-(n-1)+2=-5(-1)^n-n+3$$ $$a_{n-2}=5(-1)^{n-2}-(n-2)+2=5(-1)^n-n+4$$

$$a_{n-1}+2a_{n-2}+2n-9=-5(-1)^n-n+3+$$ $$+2\cdot 5(-1)^n-2n+8+2n-9=$$ $$=5(-1)^n-n++2=a_n$$

(iii) Given $a_n=3(-1)^n+2^n-n+2.$ Prove

$$a_n=a_{n-1}+2a_{n-2}+2n-9,n\geq2$$ $$a_{n-1}=3(-1)^{n-1}+2^{n-1}-(n-1)+2=$$ $$=-3(-1)^n+2^{n-1}-n+3$$ $$a_{n-2}=3(-1)^{n-2}+2^{n-2}-(n-2)+2=$$ $$=3(-1)^n+2^{n-2}-n+4=$$

$$a_{n-1}+2a_{n-2}+2n-9=$$ $$=-3(-1)^n+2^{n-1}-n+3+$$ $$+2\cdot3(-1)^n+2^{1+n-2}-2n+8+2n-9=$$ $$=3(-1)^n+2^n-n+2=a_n$$

(iv) Given $a_n=7\cdot 2^n-n+2.$ Prove

$$a_n=a_{n-1}+2a_{n-2}+2n-9,n\geq2$$ $$a_{n-1}=7\cdot 2^{n-1}-(n-1)+2=7\cdot 2^{n-1}-n+3$$ $$a_{n-2}=7\cdot 2^{n-2}-(n-2)+2=7\cdot 2^{n-2}-n+4$$

$$a_{n-1}+2a_{n-2}+2n-9=$$ $$=7\cdot 2^{n-1}-n+3+$$ $$+2\cdot7\cdot 2^{n-2}-2n+8+2n-9=$$ $$=7\cdot 2^n-n+2=a_n$$