1. ## Mathematical Induction

$U_{n+1} = U_n +4+4(U_n+1)^{1/2}$
$prove that U_n= 4(n+1)^2 -1$

2. Hello, helloying!

There must be typos in the problem.
The claim is not true.

$U_{n+1} \:=\:U_n +4+ 4(n+1)^{\frac{1}{2}}$ .[1]
. . . . . . . . $U_1 = 15$

Prove by induction that: . $U_n \:=\:4(4n+1)^2 -1$ .[2]

The recursion formula has a square root.
I don't see how the $n^{th}$ term could always be an integer.

In fact, we can't determine the second term of the sequence!

In [1], let $n = 1\!:\;\;U_2 \;=\;U_1 + 4 + 4\sqrt{1+1}$

. . . . . . . . . . . . . . . . $=\; 15 + 4 + 4\sqrt{2}$

. . . . . . . . . . . . . . . . $=\;19 + 4\sqrt{2}$

In [2], let $n = 2:\;\;U_2 \;=\;4(4\cdot2+1)^2-1$

. . . . . . . . . . . . . . . . $=\; 4(81) - 1$

. . . . . . . . . . . . . . . . $=\; 323$

3. Originally Posted by Soroban
Hello, helloying!

There must be typos in the problem.
The claim is not true.

The recursion formula has a square root.
I don't see how the $n^{th}$ term could always be an integer.

In fact, we can't determine the second term of the sequence!

In [1], let $n = 1\!:\;\;U_2 \;=\;U_1 + 4 + 4\sqrt{1+1}$

. . . . . . . . . . . . . . . . $=\; 15 + 4 + 4\sqrt{2}$

. . . . . . . . . . . . . . . . $=\;19 + 4\sqrt{2}$

In [2], let $n = 2:\;\;U_2 \;=\;4(4\cdot2+1)^2-1$

. . . . . . . . . . . . . . . . $=\; 4(81) - 1$

. . . . . . . . . . . . . . . . $=\; 323$
Thanks i have edited my post. sorry about the typo

4. 1. Write the claim that you need to prove in the following form: "for all n >=1, ... ". Call the part denoted by ellipses, which contains n, by P(n). This is the induction statement, or the induction hypothesis (IH). Note that for each particular n, P(n) is not a number; it is a proposition that is either true or false.

2. Prove P(1). This is the base case.

3. Prove "for all n, P(n) implies P(n+1)". For this, fix an arbitrary n and assume P(n) (that's why it is called the IH). Use it to prove P(n+1).

5. Originally Posted by helloying
$U_{n+1} = U_n +4+4(U_n+1)^{1/2}$

$Prove\ that\ U_n= 4(n+1)^2 -1$
You can set it up as follows if you like...

$\displaystyle\ U_{k+1}=U_k+4+4\left(U_k+1\right)^{\frac{1}{2}}$

Then...

P(k)

$U_k=4(k+1)^2-1$

P(k+1)

$U_{k+1}=4(k+2)^2-1$

For the proof, try to show

$U_k+4+4\left(U_k+1\right)^{\frac{1}{2}}=4(k+2)^2-1$

hence, try to show that

$\displaystyle\ [4(k+1)^2-1]+4+4[4(k+1)^2-1+1]^{\frac{1}{2}}=4(k+2)^2-1$

When you get through that, test for the first k or n.