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

2 Answers
Oct 27, 2017

#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#

Oct 27, 2017

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#