Question

Write an explicit formula for the sequence determined by this recursion formula: t{1} = 1/2; t{n} = t{n-1} + 1/n(n+1)

Answer 1

t{1} = 1/2; t{n} = t{n-1} + 1/n(n+1)

Answer 2

t{1} = 1/2 = 1/2 = 1/2
t{2} = 1/2 + 1/2(3) = 4/6 = 2/3
t[3] = 4/6 + 1/3(4) = 9/12 = 3/4
t{4} = 9/12 + 1/4(5) = 48/60 = 4/5

i see a pattern

Answer 3

\[t_{n} = \frac{n}{n+1}\]

Answer 4

ohhh, so you first find the pattern?

Answer 5

pattern finding is not the ideal way to do it; but it is a simple way alot of the times

Answer 6

there is a more analytical method which is long and drawn out

Answer 7

ohh, alright.

Answer 8

what is it?

Answer 9

it involves working the recurrsion back to a{1} and applying summations along the way to catch all the details

Answer 10

well, in this case t{1}

Answer 11

\[ \begin{align}t{n} &= t_{n-1} + \frac{1}{n(n+1)}\\\\
t_{n-1}& = t_{n-2} + \frac{1}{(n-1)((n-1)+1)}\\\\
&\text{substitute in for }t_{n-1}\\\\
t_{n}& = \left(t_{n-2} + \frac{1}{(n-1)((n-1)+1)}\right)+ \frac{1}{n(n+1)}\\\\
& = t_{n-2} + \frac{1}{(n-1)((n-1)+1)}+ \frac{1}{n(n+1)}\\\\
\end{align}\]
you continue in this substituing for lower and lower t{n-r} till you can restructure it

Answer 12

I don't understand what you just inputted.. o.o

Answer 13

\[t_{n-2}=t_{n-3}+\frac{1}{(n-1)((n-2)+1)}\]

thats because you are not familiar with the process as written, even tho you are familiar with replacing like values

Answer 14

im just rewriting the original recurrsion equation in terms of lesser and lesser terms

Answer 15

t{n-2} comes before t{n-1} which comes before t{n} .... just re writting the original in an order of terms that come before it

Answer 16

\begin{align}t{n} &= t_{n-1} + \frac{1}{n(n+1)}\\\\
t_{n-1}& = t_{n-2} + \frac{1}{(n-1)((n-1)+1)}\\\\
&\text{substitute in for }t_{n-1}\\\\
t_{n}& = \left(t_{n-2} + \frac{1}{(n-1)((n-1)+1)}\right)+ \frac{1}{n(n+1)}\\\\
& = t_{n-2} + \frac{1}{(n-1)((n-1)+1)}+ \frac{1}{n(n+1)}\\\\
t_{n-2}&=t_{n-3}+\frac{1}{(n-2)((n-2)+1)}\\\\
t_{n}& = \left(t_{n-3}+\frac{1}{(n-2)((n-1)} \right)+ \frac{1}{(n-1)((n)}+ \frac{1}{n(n+1)}\\\\
& = t_{n-3}+\frac{1}{(n-2)(n-1)}+ \frac{1}{(n-1)n}+ \frac{1}{n(n+1)}\\\\
& = t_{n-r}+\frac{1}{(n-(r-1))(n-(r-2))}+ \frac{1}{(n-(r-2))(n-(r-1))}+ ...\\\\
&\text{when r=n-1; t{n-r}=t{1}}\\\\
& = t_{1}+\frac{1}{(2)(3)}+ \frac{1}{(3)(2)}+ \frac{1}{(4)(3)} +...+ \frac{1}{(n)(n+1)}\\\\
\end{align}
the trick then becomes to find a summation of that tail end since we have defined it in terms of t{1}

Answer 17

but like i said, long and drawn out process

Answer 18

if you can find a pattern that works, that tends to be the simplest way to go