यदि `a_(1), a_(2)....a_(n)` समान्तर श्रेणी के पद है जहाँ `a_(n) gt 0` तो सिध्द कीजिए कि `(1)/(sqrt(a_(1))+sqrt(a_(2)))+(1)/(sqrt(a_(2))+sqrt(a_(3)))+...(1)/(sqrt(a_(n-1))+sqrt(a_(n)))=(n-1)/(sqrt(a_(1))+sqrt(a_(n)))`

Explanation

बायाँ पक्ष `=(1)/(sqrt(a_(1))+sqrt(a_(2)))+(1)/(sqrt(a_(2))+sqrta_(3))+(1)/(sqrt(a_(3))+sqrt(a_(4)))+....+(1)/(sqrt(a_(n-1))+sqrt(a_(n-1)))`
`=(sqrt(a_(2))-sqrt(a_(1)))/(a_(2)-a_(1))+(sqrt(a_(3))-sqrt(a_(2)))/(a_(3)-a_(2))+(sqrt(a_(4))-sqrt(a_(3)))/(a_(4)-a_(3))+...+(sqrt(a_(n))-sqrt(a_(n-1)))/(a_(n)-a_(n-1))`
`=(1)/(2)[(sqrt(a_(2))-sqrt(a_(1)))+(sqrt(a_(3))-sqrt(a_(2)))+(sqrt(a_(4))-sqrt(a_(3)))+....+(sqrt(a_(n))-sqrt(a_(n-1)))]`
`=(n-1)/((n-1)d)[sqrt(a_(n))-sqrt(a_(1))]=(n-1)/(a_(n)-a_(1))[sqrt(a_(n))-sqrt(a_(1))]=((n-1))/(sqrt(a_(n))+sqrt(a_(1)))=दायाँ पक्ष `