The sum of 1/1+root2+1/root2+root3+1/root3+root4 upto 99 terms
Answers
Answered by
2
Answer:
1/(sqrt1+ sqrt2) + 1/(sqrt2 + sqrt3) + ....+ 1/(sqrt99+ sqrt100)
We know that:
1/(sqrta + sqrt(1+a) = sqrt(a+1) -sqrta
==>sqrt2 - sqrt1 + sqrt3 -sqrt2 + ...+ sqrt100 -sqrt99
= -sqrt1 + sqrt100
= -1 + 10
= 9
Mark me as brainliest
Similar questions