Question 178268
When trying to simplify or expand complicated expressions involving roots, you can try substitutions to get the expression into a cleaner form (it's not required, but I find that it helps you see the big picture).



So if we let {{{x=sqrt(n+1)}}} and {{{y=sqrt(n)}}}, we can simplify



{{{(sqrt(n+1)+sqrt(n))(sqrt(n+1)-sqrt(n))}}}



to get



{{{(x+y)(x-y)}}} (note: I just replaced every {{{sqrt(n+1)}}} with "x" and every {{{sqrt(n)}}} with "y")



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Now let's FOIL {{{(x+y)(x-y)}}}



Remember, when you FOIL an expression, you follow this procedure:



{{{(highlight(x)+y)(highlight(x)-y)}}} Multiply the <font color="red">F</font>irst terms:{{{(x)*(x)=x^2}}}.



{{{(highlight(x)+y)(x+highlight(-y))}}} Multiply the <font color="red">O</font>uter terms:{{{(x)*(-y)=-x*y}}}.



{{{(x+highlight(y))(highlight(x)-y)}}} Multiply the <font color="red">I</font>nner terms:{{{(y)*(x)=x*y}}}.



{{{(x+highlight(y))(x+highlight(-y))}}} Multiply the <font color="red">L</font>ast terms:{{{(y)*(-y)=-y^2}}}.



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{{{x^2-x*y+x*y-y^2}}} Now collect every term to make a single expression.



{{{x^2-y^2}}} Now combine like terms.



So {{{(x+y)(x-y)}}} FOILs to {{{x^2-y^2}}}.



In other words, {{{(x+y)(x-y)=x^2-y^2}}}.



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{{{(x+y)(x-y)=x^2-y^2}}} Start with the last equation



{{{(highlight(sqrt(n+1))+y)(highlight(sqrt(n+1))-y)=(highlight(sqrt(n+1)))^2-y^2}}} Plug in {{{x=sqrt(n+1)}}} (ie replace every "x" with {{{sqrt(n+1)}}})



{{{(sqrt(n+1)+highlight(sqrt(n)))(sqrt(n+1)-highlight(sqrt(n)))=(sqrt(n+1))^2-(highlight(sqrt(n)))^2}}} Plug in {{{y=sqrt(n)}}} (ie replace every "y" with {{{sqrt(n)}}})



{{{(sqrt(n+1)+sqrt(n))(sqrt(n+1)-sqrt(n))=(sqrt(n+1))^2-(sqrt(n))^2}}} Simplify



{{{(sqrt(n+1)+sqrt(n))(sqrt(n+1)-sqrt(n))=n+1-(sqrt(n))^2}}} Square {{{sqrt(n+1)}}} to get {{{n+1}}} (note: the square undoes the square root)



{{{(sqrt(n+1)+sqrt(n))(sqrt(n+1)-sqrt(n))=n+1-n}}} Square {{{sqrt(n)}}} to get {{{n}}} (note: the square undoes the square root)



{{{(sqrt(n+1)+sqrt(n))(sqrt(n+1)-sqrt(n))=1}}} Combine like terms.




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Answer:



So {{{(sqrt(n+1)+sqrt(n))(sqrt(n+1)-sqrt(n))=1}}}