Question
Given the sequence 10001,100010001,1000100010001,..., prove each term of the sequence is a composite number
Given the sequence 10001,100010001,1000100010001,..., prove each term of the sequence is a composite number
a_1 = 10001 =10^{4\times1}+1= 73\times 137
a_2=100010001 = 10^{4\times2}+10^{4\times1}+1
a_3 = 1000100010001 =10^{4\times3}+ 10^{4\times2}+10^{4\times2}+10^{4\times1}+1
...
a_n = \underbrace{100010001\dots 0001}_{\text{n 0001}}= 10^{4n}+10^{4n-1}+\dots+10^4+1
Using the Sum formula for a geometric sequence
S_n=\dfrac{a_1(1-r^n)}{1-r}
a_n = \dfrac{1-10^{4(n+1)}}{1-10^4} =\dfrac{1-(10^{2n+2})^2}{1-10^4} =\dfrac{(10^{2n+2}-1)(10^{2n+2}+1) }{10^4-1}
Since 10^4-1< 10^{2n+2}-1 < 10^{2n+2}+1
a_n is not a prime number but a composite number