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How do you calculate 1/1*2*3*4 + 1/2*3*4*5... 1/97*98*99*100?

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parama9000 | Student, Grade 11 | (Level 1) Valedictorian

Posted September 14, 2013 at 2:41 PM via web

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How do you calculate 1/1*2*3*4 + 1/2*3*4*5... 1/97*98*99*100?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 14, 2013 at 3:09 PM (Answer #1)

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The sum `1/(1*2*3*4) + 1/(2*3*4*5)... 1/(97*98*99*100)` has to be determined.

Looking at each of the term, we see:

`1/(1*2*3*4)`

= `(1/3)*(1/(1*2*3))`

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

`1/(2*3*4*5)`

= `(1/3)*(1/(2*4*5))`

= `(1/3)*(1/(2*3*4) - 1/(3*4*5))`

...

`1/(97*98*99*100)`

= `(1/3)*(1/(97*98*99) - 1/(98*99*100))`

Using this: `1/(1*2*3*4) + 1/(2*3*4*5) + ... + 1/(97*98*99*100)`

= `(1/3)*(1/(1*2*3) - 1/(2*3*4) + 1/(2*3*4) - 1/(3*4*5)+ ... -1/(98*99*100))`

= `(1/3)*(1/(1*2*3) -1/(98*99*100))`

= `(1/3)(1/6 - 1/970200)`

= `970194/17463600`

`~~ 0.05555`

The required sum is `970194/17463600`

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