Question

A sequence {an} is generated by the recursive formulas a1 = 5 and an = an-1 + 5(-1)n. Find, a343, the 343rd term of the sequence

Answer 1

$a_1 = 5\ \ \ and \ \ \ a_n = a_{n-1} + 5\cdot(-1)^n\\\\(-1)^{even\ number}=1\ \ \ and\ \ \ (-1)^{odd\ number}=-1\\\\a_2=a_1+5\cdot(-1)^2=5+5=10\\\\ a_3=a_2+5\cdot(-1)^3=10-5=5=a_1\\\\a_4=a_3+5\cdot(-1)^4=5+5=10=a_2\\\\\\if\ n\ \rightarrow\ even\ number\ \ \ \Rightarrow\ \ \ a_n=10\\if\ n\ \rightarrow\ odd\ number\ \ \ \ \ \Rightarrow\ \ \ a_n=5\\\\\Rightarrow\ \ \ a_{343}=5$

Answer 2

$a_n=a_{n-1}+5(-1)^n\\\\if\ n\ is\ even\ then:a_n=a_{n-1}+5(-1)^n=a_{n-1}+5\\\\if\ n\ is\ odd\ then:a_n=a_{n-1}+5(-1)^n=a_{n-1}-5\\\\a_1=5\\a_2=5+5=10\\a_3=10-5=5\\a_4=5+5=10\\a_5=10-5=5\\\vdots\\a_{343}=5\ \ \ \ (343\ is\ odd\ number)$