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

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### 3 Answers

To find the product of (1-1/4)(1-1/9) (1-1/16)...(1-1/100)

We rewrite the product as:

(1+1/2)(1-1/2) } { (1+1/3)(1-1/3)} {1+1/4)(1-1/4)}...

(3/2)(1/2)}{ 4/3)(2/3)}. .... {(11/100 )(9/10)}

Rearrange as

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

= 11/2*1/10

= 11/20

Therefore the give product = 11/20.

We notice that the factors 1/4,1/9,...1/100 could be written as:

1/2^2, 1/3^2,...,1/10^2

For calculating the product, we notice that in the brackets we have differences of squares, type a^2 – b^2= (a-b)(a+b)

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

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

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

Now, we’ll do the arithmetical operations in each paranthesis

(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:

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

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

**P=(1/2)*(11/10)=11/20**

We need to find the value of the product as given by (1-1/4)(1-1/9)...(1-1/100)

Now (1 - 1/4)(1 - 1/9)...(1 - 1/100)

We see that the denominator of consecutive terms is the square of a number increasing by 1 and it starts with 2 and the numerator is one less than the denominator.

=> (3/4)(8/9)...(99/100)

We take all the terms that we have to include in finding the product.

=> (3*8*15*24*35*48*63*80*99) / (4*9*16*25*36*49*64*81*100)

cancel common terms

=> 11/20

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