# Show that number n=`sum_1^99` 1/(`sqrt k` + `sqrt(k+1)` ) is natural number?

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### 1 Answer

The general term of summation is given by the fraction `1/(sqrt k + sqrt(k+1))` , hence, you should rationalize the denominator to eliminate the square roots, such that:

`1/(sqrt k + sqrt(k+1)) = (sqrt k - sqrt(k+1))/((sqrt k + sqrt(k+1))(sqrt k - sqrt(k+1)))`

Converting the product `((sqrt k + sqrt(k+1))(sqrt k - sqrt(k+1))) ` into a difference of squares, yields:

`((sqrt k + sqrt(k+1))(sqrt k - sqrt(k+1))) = k - (k + 1)`

`((sqrt k + sqrt(k+1))(sqrt k - sqrt(k+1))) = k - k - 1 = -1`

`1/(sqrt k + sqrt(k+1)) = -(sqrt k - sqrt(k+1))`

`1/(sqrt k + sqrt(k+1)) = sqrt(k+1) - sqrt k`

Hence, you need to evaluate the given summation, such that:

`sum_(k=1)^99 1/(sqrt k + sqrt(k+1)) = sum_(k=1)^99 sqrt(k+1) - sqrt k`

`sum_(k=1)^99 (sqrt(k+1) - sqrt k) = sqrt(1+1) - sqrt 1 + sqrt(2+1) - sqrt 2 + sqrt(3+1) - sqrt 3 + .... + sqrt(98+1) - sqrt 98 + sqrt(99+1) - sqrt 99`

`sum_(k=1)^99 sqrt(k+1) - sqrt k = sqrt 2 - 1 + sqrt3 - sqrt2 + sqrt4 - sqrt3 + ... + sqrt99 - sqrt98 + sqrt100 - sqrt99`

Reducing duplicate terms yields:

`sum_(k=1)^99 sqrt(k+1) - sqrt k = sqrt100 - 1`

`sum_(k=1)^99 sqrt(k+1) - sqrt k = 10 - 1`

`sum_(k=1)^99 sqrt(k+1) - sqrt k = 9`

**Hence, evaluating the given number `n = sum_(k=1)^99 sqrt(k+1) - sqrt k` yields `n = 9` is a natural number.**