Here are four terms of a sequence of functions 1/1 2/3 3/5 4/7

Question

Here are four terms of a sequence of functions

1/1 2/3 3/5 4/7

a- write down the expression of term of n

b- using algebra prove that when the square of any odd number is divided by 4 the remainder is 1

Answer 1

as number increases by 1, the denominator increases by 2

a1 = 1/1 = 1

a2 = 2/3 = .666...

a3 = 3/5 = .600...

a4 = 4/7 = .571428571428571428...

a5 = 5/9 = .555...

the values decrease each term from 1 to (1+n)/(1+2n) for the nth term

as n approaches infinity an approaches 1/2 = .5000.. = .5

the nth term

=an = 1-(1+n-2)/(1+2n-2)

an = 1-(n-1)/(2n-1)

a1 =1 - (0/-1) =1

a2 = 1-(2-1)/(4-1) = 1-1/3 = 2/3

a3 = 1-(3-1)/(6-1) = 1-2/5 = 3/5

a4 = 1-(4-1)/(8-1) = 1- 3/7 = 4/7

a5 = 1-(5-1)/(10-1) = 1-4/9 = 5/9 = .555...

a101 = 1 -(101-1)/(202-1) = 1-100/201 = 101/201 = about 0.502

an = ever closer to .5 as n increases

a2 = 1+