# Evaluate the product (1 - 4^-1)(1 - 9^-1)...(1 - 100^-1).

*print*Print*list*Cite

### 2 Answers

To find the product (1 - 4^-1)(1 - 9^-1)...(1 - 100^-1).

We rewrite the product as

(1-1/4)(1-1/9)(1/16)(1-1/25)....(1-1/100)

(1-1/2^2)(1-1/3^2)(1-1/4^2) ....(1-1/10^2).....(1)

Consider 1-1/n^2 = (1-1/n)(1+1/n) .

Therefore (1-1/2^2) = (1-1/2)(1+1/2) = (1/2)(3/2)

1-1/3^2 = (2/3)(4/3).

(1-1/2^2)(1-1/3^2)(1-/3^2) ...(1-1/10^2) = {(1/2)(3/2)][(2/3)(4/3)][(3/4)(5/4)]....(9/10)(11/10)}

= {(1/2)(2/3)(3/4)....(9/10)}{((3/2)(4/3)(5/4)...(11/10)}

= {1/10} {11/2}= 11/20

Therefore (1 - 4^-1)(1 - 9^-1)...(1 - 100^-1) = 11/20 = 0.55.

For calculating the product, first we'll have to re-write the terms from each brackets, using the rule of negative exponent:

a^-b = 1/a^b

P = (1 - 1/4)(1 - 1/9)...(1 - 1/10)

We notice that each factor of the product is a differences of squares:

We'll write 1/4 = 1/2^2

So, (1- 1/2^2)=(1-1/2)(1+1/2)

We'll write 1/9 = 1/3^2

(1- 1/3^2)=(1-1/3)(1+1/3) and so on till

.............................................

We'll write 1/100 = 1/10^2

(1-1/10^2)=(1-1/10)(1+1/10)

Now, we’ll compute each factor:

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

(1-1/3)(1+1/3)=(2/3)*(4/3)

…………………………….

(1-1/10)(1+1/10)=(9/10)*(11/10)

We’ll re-group the factors in a convenient way:

P = [(1/2)*(3/2)]*[(2/3)*(4/3)]*….*[ (9/10)*(11/10)]

P = (1/2)*[ (3/2)*(2/3)]*…*[(10/9)*( 9/10)*(11/10)

P = (1/2)*(11/10)

P = **11/20**