# Find a formula for f(n), assuming the pattern continues a) -1/4, 2/9, -3/16, 4/25... b) 1, 2/3, 3/5, 4/7, 5/9... c) 1,2,1/3,4,1/5,6...

hala718 | Certified Educator

a) -1/4 , 2/9, -3/16, 4/25

We notice that the numerator = n*[(-1)^n]   ( n= 1, 2, 3, ...)

We also notice that the denominator = (n+1)^2   (n= 1, 2, ...)

Then f(n) = n*(-1)^n/ (n+1)^2

b) 1, 2/3, 3/5, 4/7, 5/9...

The numerator = n   (n=1,2,3...)

The denominator = (2n-1)

Then , f(n) = n/(2n-1)    (n= 1, 2, 3,,,,)

c) 1,2,1/3,4,1/5,6...

Let us rewrite:

1, 2, 3^-1, 4, 5^-1, 6, 7^-1, .....

Then, f(n) = n^[(-1)^n]     (n= 1, 2, 3, ....)

kjcdb8er | Certified Educator

a) -1/4, 2/9, -3/16, 4/25...

(-1)^n * n/(n+1)^2

b) 1, 2/3, 3/5, 4/7, 5/9...

n/( 2n-1)

neela | Student

To find the formula for f(n)

a) -1/4, 2/9, -3/16, 4/25...

f(1) = ( 1)^1*1 /((1+1)^2 = 1/4,

f(2) = (-1)^2*2((2+1)^2 = 2/9,

f(3) = (-1)^3*3/ (3+1)^2 = -3/16,

f(4)= (-1)^4*4/(4+1)^2 =  4/25

So f(n) = (-1)^n *n/(n+1)^2., n =1,2,3....

b) 1, 2/3, 3/5, 4/7, 5/9...

f(1)  = 1/2*1-1)

f(2) = 2/(2*2-1) = 2/3

f(3) = 3/(3*2-1) = 3/5

f(4) = 4/(4*2-1) = 4/7

f(5) = 5/(5*2-1) = 5/9

f(n) = n/(n*2-1) = n/(2n-1).

c) 1,2,1/3,4,1/5,6...

f(1) =1/1 =1

f(2) =2

f(3) = 1/3

f(4) = 4

f(5) = 1/5

f(6) = 6.

f(n) = 1/n  if n is odd . f(n) = n if n is even.