Evaluate: (1 - 1/10)(1 - 1/11)(1 - 1/12)...(1 - 1/99)(1 - 1/100)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find (1 - 1/10)(1 - 1/11)(1 - 1/12)...(1 - 1/99)(1 - 1/100).

Now (1 - 1/10)(1 - 1/11)(1 - 1/12)...(1 - 1/99)(1 - 1/100).

=> [(10 - 1)/10][(11 - 1)/11][(12 - 1)/12]...[(99-1)/99][(100-1)/100].

=> (9/10)(10/11)(12/11)...(98/99)(99/100)

We see that all terms get cancelled except 9/100.

Therefore (1 - 1/10)(1 - 1/11)(1 - 1/12)...(1 - 1/99)(1 - 1/100) = 9/100.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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Let us assume that the number is E.

==> E = (1-1/10) ( 1-1/11) ( 1- 1/12) .... (1-1/99)( 1- 1/100)

Let us rewrite the terms by using the common denominator.

==> 1- 1/10 = 10/10 - 1/10 = 9/10

==> 1- 1/11 = 11/11 - 1/11 = 10/11

==> 1- 1/12 = 12/12 - 1/12 = 11/12

.........

==> 1- 1/99 = 99/99 - 1/99 = 98/99

==> 1 - 1/100 = 100/100 - 1/100 = 99/100

Now we will rewrite the terms.

==> E = 9/10 * 10/11 * 11/12 ............ * 98/99 * 99/100

After we reduce similar terms.

==> E = 9 / 100

Then the number ( 1- 1/10)(1-1/11) ....(1-1/100) = 9/100

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neela | High School Teacher | (Level 3) Valedictorian

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Q:  (1 - 1/10)(1 - 1/11)(1 - 1/12)...(1 - 1/99)(1 - 1/100).

A:

Let x = (1-1/10)(1-1/11)(1-1/12)...(1-1/100).

=> x = {(10-1)/10)}{(11-1)/11}{(12-1)/12}...{(100-1)/100}

=> x = (9/10)(10/11)(11/12)(12/13)....(99/100)

=> x= (9*10*11*12*...99)/(10*11*12*13*...100)

=> x = 9* {(10*11*12.13..99)/(10*11*12.13*...99)}/100

=> x = 9/100 = 0.09.

Therefore x = 0.09.

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