Telescoping Square Roots Galore

Algebra Level 3

Evaluate

1 n 1 n + 1 . \frac{1}{\sqrt{n} } - \frac{1}{ \sqrt{n+1} }.

1 n ( n + 1 ) \frac{ 1}{ \sqrt{ n (n+1) } } 1 n × ( n + 1 ) + n + 1 × n \frac{ 1} { \sqrt{n} \times (n+1) + \sqrt{n+1} \times n } 1 n + 1 n \frac{ 1} { \sqrt{n+1} - \sqrt{n} } n + 1 n \sqrt{ n+1 } - \sqrt{n}

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2 solutions

Chung Kevin
Apr 10, 2014

This was tricky, because it's hard to see how the denominator changed so drastically. One would normally expect that the denominator was just n × n + 1 \sqrt{n} \times \sqrt{n+1} , but this is not the case.

First, we combine fractions to get

n + 1 n n × n + 1 . \frac{ \sqrt{n+1} - \sqrt{n} } { \sqrt{n} \times \sqrt{n+1} }.

Notice that this is not any of the options. We rationalize the numerator, by multiplying it with n + 1 + n \sqrt{ n+1} + \sqrt{n} , which gives us

n + 1 n n × n + 1 × n + 1 + n n + 1 + n = 1 n × n + 1 × ( n + 1 + n ) . \frac{ \sqrt{n+1} - \sqrt{n} } { \sqrt{n} \times \sqrt{n+1} } \times \frac { \sqrt{ n+1} + \sqrt{n} } { \sqrt{ n+1} + \sqrt{n} } = \frac{ 1 } { \sqrt{n} \times \sqrt{n+1} \times ( \sqrt{ n+1} + \sqrt{n} ) }.

Finally, we expand and multiply, to obtain

1 n × ( n + 1 ) + n + 1 × n . \frac{1} { \sqrt{ n} \times (n+1) + \sqrt{n+1} \times n } .

just do the normal subtraction and then rationalize the numerator.

Write 1 of N as (n+1) - n and proceed.

Ravindra Pradhan - 7 years, 2 months ago

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