[administration]

Q. The sum of the series:
1/(root2+root1) + 1/(root2+root3) +……… 1/(root120+root121)

(a)10
(b)11
(c)12
(d)None

asked by Kanupriya Verma

[Wonder Boy]

1/(root2+root1) = (root2- root1)/(2-1) = root2-root1

so 1/(root2+root1) + 1/(root2+root3) +……… 1/(root120+root121)

= (root2 – root1) + (root3 – root2) + (root4 – root3)……… + (root121 – root120)

= (root2 – root1) + (root3 – root2) + (root4 – root3)……… + (root121 – root120)

= root 121 – root 1 = 11-1 = 10 Ans

[Wonder Boy]

In such problems having roots in denominator just rationalise .. u will find terms automatically arranged in such a manner that u can cancel max of them leaving 2 or 4 terms.. if there is no root then expressing the terms as difference of two may be a bit difficult depending on question

[Kanupriya Verma]

thanks…

[Author]

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