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prove $(\sqrt x+\sqrt {x+1})^n$ where n is an odd natural number can always be written as $\sqrt y + \sqrt{y+1}$ while y is also a natural number .

I have tried using induction by proving the question when n is an even integer which is easy,then I tried reducing $(\sqrt x+\sqrt {x+1})^n$ to $(\sqrt z+\sqrt {z+1})$.$(\sqrt x+\sqrt {x+1})$ in my induction process but I hit a hard wall here and don't seem to figure out a way to write these two parentheses in form of a unified $\sqrt y + \sqrt{y+1}$ .

Should I change my approach?

Is there any other way to prove this ?

s1234s1234d95829
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    $n$ is odd so you don't increase it by $1$ but by $2$. – kingW3 Jun 11 '21 at 12:32
  • I might have made a mistake trying to prove the relation without strong induction but if you accept the theorem for every positive even integer (by proving it using induction) you should be able to manipulate the resulting new terms to $\sqrt y + \sqrt{y+1}$ where I can't , I appreciate any other methods even without induction – s1234s1234d95829 Jun 11 '21 at 12:35

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