Show that for all `n in N`.
`sqrt(a+sqrt(a+sqrt(a+....+sqrt(a))))lt (1+sqrt((4a+1)))/(2)`
where'a' is fixed positive number and n radical signs are taken on LHS.

Explanation

Let `P(n):sqrt(a+sqrt(a+sqrt(a+.....+sqrt(a))))lt 1+sqrt(((4a+1))/(2)`
Step I For `n=1` , then `sqrt(a)lt 1+sqrt((4a+1))/(2)`
`rArr 2sqrt(a) lt 1+sqrt((4a+1))`
`rArr 4alt 1+4a+1+2sqrt((4a+1))`
`rArr 2sqrt((4a+1))+2 gt 0` which is true .
Therefore , P(1) is ture .
Step II Assume it is true for n`=k` .
`P(k): ubrace(sqrt(a+sqrt(a+sqrt(a+....+sqrt(a)))))_("k radical signs")lt (1+sqrt(4a+1))/(2)`
Step III For `n =k+1`,
`P(k+1):ubrace(sqrt(a+sqrt(a+sqrt(a+....+sqrt(a)))))_("(k+1) radical signs")lt (1+sqrt((4a+1)))/(2)`
From assumption step
`ubracesqrt(a+sqrt(a+sqrt(a+......+sqrt(a))))_("k radical signs")lt (1+sqrt(4a+1))/(2)`
`a+ubracesqrt(a+sqrt(a+sqrt(a+......+sqrt(a))))_("k radical signs")lt (1+sqrt((4a+1)))/(2)`
`rArrubracesqrt(a+sqrt(a+sqrt(+sqrt(a+....+sqrt(a)))))_((k+1)"radical signa")lt sqrt(a+(1+sqrt((4a+1)))/(2))`
`=sqrt((2a+1+sqrt((4a+1)))/(2))=sqrt((4a+2+2sqrt((4a+1)))/(4))`
`=sqrt(((sqrt((4a+1)))^2+1+2sqrt((4a+1)))/(4))`
`=sqrt(((1+sqrt((4a+1)))/(2))^2)=(1+sqrt((4a+1)))/(2)`
`therefore ubracesqrt(a+sqrta+sqrt(a+sqrt(a+....+sqrt(a))))_((k+1)"radical signs")lt (1+sqrt((4a+1)))/(2)`
which is true for `n=k+1`.
Hence , by the principle of mathematical induction , the result is true for all ` n in N `.