Sum of a sequence?

Let $\displaystyle s_n = \sum_{k=1}^n a_k$ . Then we have:

$\begin{aligned} s_{n+1} & = \frac 12 \left(a_{n+1} + \frac 1{a_{n+1}}\right) & \small \color{#3D99F6} \text{Note that }s_{n+1} = s_n + a_{n+1} \\ 2 (s_n + a_{n+1}) & = a_{n+1} + \frac 1{a_{n+1}} & \small \color{#3D99F6} \text{Rearranging} \\ a_{n+1}^2 + 2s_na_{n+1} - 1 & = 0 & \small \color{#3D99F6} \text{Solving the quadratic for }a_{n+1} \\ a_{n+1} & = \sqrt{s_n^2 +1} - s_n \\ s_{n+1} - s_n & = \sqrt{s_n^2 +1} - s_n \\ s_{n+1} & = \sqrt{s_n^2 +1} \\ s_{n+1}^2 & = s_n^2 +1 \end{aligned}$

Since $s_1=a_1$ , we have $a_1 = \dfrac 12 \left(a_1 + \dfrac 1{a_1}\right)$ $\implies a_1 = 1 = s_1$ . Then we have:

$\begin{aligned} s_2^2 & = s_1^2 + 1 = 2 \\ s_3^2 & = s_2^2 + 1 = 3 \\ s_4^2 & = s_3^2 + 1 = 4 \\ \implies s_{100}^2 & = 100 & \small \color{#3D99F6} \text{Since } s_n = \sum_{k=1}^n a_k \\ \sum_{k=1}^{100} a_k & = \sqrt{100} = \boxed{10} \end{aligned}$

Subtracting $a_n$ from both sides of the definition equation gives $\frac12 \left( \frac{1}{a_n} - a_n \right) = \frac12 \left( \frac{1}{a_{n-1}} + a_{n-1} \right)$ . This is an easier formulation to work with.

A bit of experimenting with the first few terms suggests that $a_n=\sqrt{n}-\sqrt{n-1}$ , and indeed this is easy to prove by induction.

The sum of these terms telescopes, and we get

$\sum_{k=1}^{100}a_k=\boxed{10}$

Let $S_n = \displaystyle \sum_{k = 1}^{n} a_k$ . Then $S_n = S_{n - 1} + a_n = \frac{1}{2}(a_n + \frac{1}{a_n})$ , which using the quadratic equation on $a_n$ on the right side solves to $a_n = -S_{n - 1} + \sqrt{S_{n - 1}^2 +1}$ , so that $S_n = \sqrt{S_{n - 1}^2 + 1}$ . This can inductively be shown to be $S_n = \sqrt{n}$ , so that $\displaystyle \sum_{k = 1}^{100} a_k = S_{100} = \sqrt{100} = \boxed{10}$ .

$\sum_{k=1}^{n}{a_k}=\frac{1}{2}\left(a_n+\frac{1}{a_n}\right) \to \sum_{k=1}^{n-1}\left({a_k}\right)+a_n =\frac{1}{2}\left(a_n+\frac{1}{a_n}\right) \to \frac{1}{2}\left(a_{n-1}+\frac{1}{a_{n-1}}\right) +a_n = \frac{1}{2}\left(a_n+\frac{1}{a_n}\right) \\\to a_{n-1}+\frac{1}{a_{n-1}} =-a_n+\frac{1}{a_n} \to \left(a_{n-1}+\frac{1}{a_{n-1}}\right)^2 = \left(-a_n+\frac{1}{a_n}\right) ^2 \to \left(a_{n-1}+\frac{1}{a_{n-1}}\right)^2 = \left(a_n+\frac{1}{a_n}\right) ^2 -4$ notice we used $\left( -x+\dfrac{1}{x}\right)^2 = \left( x+\dfrac{1}{x}\right)^2 -4$ . let $b_n = \left(a_n+\frac{1}{a_n}\right) ^2$ , then we have $b_{n-1} = b_{n}-4 \to b_{n} = b_1 +4(n-1) = 4n$ we used the fact that $b_n$ forms an arithmetic sequence, and that $b_1$ is found by plugging in $n=1$ into the equation given in the question, i.e $a_1 = \frac{1}{2}\left(a_1+\frac{1}{a_1}\right) \to a_1 = 1 \to b_1 = \left(a_1+\frac{1}{a_1}\right) ^2 = 4$ note only the positive solution was taken into account, as $a_n$ is a positive sequence.

hence we have $\sum_{k=1}^{n}{a_k}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right) = \dfrac{\sqrt{b_{n}}}{2} = \dfrac{\sqrt{4n}}{2} = \sqrt{n}$ plugging in $n=100$ we will have $\sqrt{100} = \boxed{10}$ . also by solving $\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right) =\sqrt{n} \to a_n = \sqrt{n}-\sqrt{n-1}$ where again only the positive solution was taken, proving the existence of such a sequence.