这是一份northeastern东北大学(美国) MATH 1120作业代写的成功案
问题 1.
$$
a_{k+1}>a_{k}
$$
so
$$
a_{k+1}+6>a_{k}+6
$$
证明 .
and
$$
\frac{1}{2}\left(a_{k+1}+6\right)>\frac{1}{2}\left(a_{k}+6\right)
$$
Thus
$$
a_{k+2}>a_{k+1}
$$
MATH 1120COURSE NOTES ：
Let $a$ and $b$ be positive numbers with $a>b$. Let $a_{1}$ be their arithmetic mean and $b_{1}$ their geometric mean:
$$
a_{1}=\frac{a+b}{2} \quad b_{1}=\sqrt{a b}
$$
Repeat this process so that, in general,
$$
a_{n+1}=\frac{a_{n}+b_{n}}{2} \quad b_{n+1}=\sqrt{a_{n} b_{n}}
$$
(a) Use mathematical induction to show that
$$
a_{n}>a_{n+1}>b_{n+1}>b_{n}
$$(c) Show that $\lim {n \rightarrow \infty} a{n}=\lim {n \rightarrow \infty} b{n}$. Gauss called the common value of these limits the arithmetic-geometric mean of the numbers $a$ and $b$.