Use the principle of mathematical induction to prove that for all `n in N`
`sqrt(2+sqrt(2+sqrt2+...+...+sqrt2))=2cos ((pi)/(2^(n+1)))`
When the LHS contains n radical signs.

Explanation

Let`P(n)=sqrt(2+sqrt(2+sqrt(2+....+....+sqrt(2))))`
`=2 cos ((pi)/(2^(n+1)))` .......(i)
Step I For `n=1`.
LHS of Eq. `(i) =sqrt(2) ` and RHS Eq. (i)` =2 cos ((pi)/(2^2))`
`=2cos ((pi)/(4))=2.(1)/(sqrt(2))=sqrt(2)`
Therefore , P(1) is true.
Step II Assume it is true for `n=k`,
`P(k)=ubrace(sqrt(2+sqrt(2+sqrt(2+....+....+sqrt(2)))))_("k radical sign")=2cos((pi)/(2^(k+1)))`
Step III For `n=k+1` ,
`therefore P(k+1)=ubrace(sqrt(2+sqrt(2+sqrt(2+....+....+sqrt(2)))))_("(k+1) radical sign")`
`=sqrt({2+P(k)})`
`=sqrt(2+2cos. ((pi)/(2^k+1)))`
`=sqrt(2(1+cos.((pi)/(2^(k+1)))))`
`=sqrt(2(1+2cos^2((pi)/(2^(k+2)))-1))`
`=sqrt(4 cos^2((pi)/(2^(k+2)))=2 cos ((pi)/(2^(k+1))))`
This shows that the result is true for `n=k+1`. Hence , by the principle of mathematical induction , the result is true for all `n in N`.