Question

Simplify `1/15+1/35+1/63+1/99+1/143`?

Answer 1

`1/15+1/35+1/63+1/99+1/143=5/39`

Explanation:

`1/15+1/35+1/63+1/99+1/143`

= `1/(3xx5)+1/(5xx7)+1/(7xx9)+1/(9xx11)+1/(11xx13)`

= `(7xx9xx11xx13+3xx9xx11xx13+3xx5xx11xx13+3xx5xx7xx13+3xx5xx7xx9)/(3xx5xx7xx9xx11xx13)`

= `(9009+3861+2145+1365+945)/135135`

= `17325/135135`

= `3465/27027` - dividing by `5`

= `385/3003` - dividing ny `9`

= `35/273` - dividing by `11`

= `5/39` - dividing by `7`

Answer 2

5/39

Explanation:

Given, `1/15+1/35+1/63+1/99+1/143`

First of all multiply Numerator and Denumerator by 2/2 and we get,

`rArr 2/2(1/15+1/35+1/63+1/99+1/143)`

`rArr 1/2(2/15+2/35+2/63+2/99+2/143)`

`rArr 1/2[(1/3-1/5)+(1/5-1/7)+(1/7-1/9)+(1/9-1/11)+(1/11-1/13)]`

`rArr 1/2[1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13]`

`rArr 1/2[1/3-1/13]`

`rArr 1/2[(13-3)/39]`

`rArr 1/2. 10/39`

`rArr 5/39`