Italy
^@ \dfrac{ 1 } { 1 + \sqrt{2} } + \dfrac{ 1 } { \sqrt{2}+\sqrt{3} } + \dfrac{ 1 } { \sqrt{3}+\sqrt{4 } } + \ldots + \dfrac{ 1 } { \sqrt{ 15 } + \sqrt{ 16 } } = ? \space\space\space\space\space^@
Answer:
^@ 3 ^@
- If we multiply the numerator and denominator of the first term by ^@ \sqrt{ 2 } - 1 ^@ we get:
^@ \begin{align} \dfrac{ 1 } { 1 + \sqrt{2} } &= \dfrac{ ( \sqrt{2} - 1 ) } { ( \sqrt{2} + 1 ) ( \sqrt{2} - 1 ) } \\ &= \dfrac{ \sqrt{2} - 1 } { 2 - 1 } && [\text{Since, } a^2 - b^2 = (a + b)(a - b)] \\ &= \sqrt{2} - 1 \\ \end{align} ^@ - Similarly, if we multiply numerator and denominator of the second term by ^@ \sqrt{ 3 } - \sqrt{ 2 } ^@, we get:
^@ \begin{align} \dfrac{ 1 } { \sqrt{2}+\sqrt{3} } &= \dfrac{ \sqrt{3} - \sqrt{2} } { ( \sqrt{3}+\sqrt{2} ) ( \sqrt{3} - \sqrt{2} ) } \\ &= \dfrac{ \sqrt{3} - \sqrt{2} } { 3 - 2 } && [\text{Since, } a^2 - b^2 = (a + b)(a - b)] \space\space\space\space\space \\ &= \sqrt{3} - \sqrt{2} \end{align} ^@ - Using the same technique with the other terms too, we can re-write the expression in question as:
^@ \begin{align} & \dfrac{ 1 } { 1 + \sqrt{2} } + \dfrac{ 1 } { \sqrt{2}+\sqrt{3} } + \dfrac{ 1 } { \sqrt{3}+\sqrt{4 } } + \ldots + \dfrac{ 1 } { \sqrt{ 15 } + \sqrt{ 16 } } \\ = & (\sqrt{ 2 } - 1 ) + (\sqrt{ 3 } - \sqrt{ 2 } ) + (\sqrt{ 4 } - \sqrt{ 3 } ) + \ldots + (\sqrt{ 16 } - \sqrt{ 15 } ) \\ = & -1 + \sqrt{ 16 } \\ = & -1 + 4 \\ = & 3 \end{align} ^@
