Question

If sequence an (aka a sub n) is defined as an=sqrt(2+a(n-1)), and a1=sqrt(2), prove by induction that an is increasing.

Answer 1

@ikram002p

Answer 2

\(\large a_n^2 = 2 + a_{n-1}\)

Answer 3

wait why u need induction to show that its increasing ?
just show that a_n <a_n+1
which is the same
\((a_n)^2<a_{n+1}^2\)

Answer 4

ok lets induction since no response :P

Answer 5

if you show that for the general case, that is essentially induction.

Answer 6

for
first show for \((a_1)^2 =2\)
\((a_2)^2=2+\sqrt2\)
\(a_1^2<a_2^2 \) hold

then assume its true for k
\(a_k^2<a_{k+1}^2\)

show for
\(a_{k+1}^2<a_{k+2}^2\)

ugh dont feel like continue :3 but its smothly could be done bbye

Answer 7

It looks good, but I already got up to that part.

Answer 8

the next step is what I was stuck on

Answer 9

\(\large\begin{align} a_{k+2}^2 &= 2 +a_{k+1} \\&\gt 2+a_k \\&=a_{k+1}^2 \end{align}\)

Answer 10

So, if \(\large a_k^2 \lt a_{k+1}^2\) holds, so does \(\large a_{k+1}^2 \lt a_{k+2}^2\)

that ends the induction proof

Answer 11

thanks