Series/Sequences of numbers

[Rapper]

Give an explicit form each of the terms:

 

a(1) = 1

a(2) = 3

a(n+1) = [a(n) + a(n+1)]/2

 

I just started with this topic. How can I calculate what a(n) is?

Should I replace a(n) by 1 and a(n+1) by 3 and divide the result by 2? Would that give me de?

[mathematic]

You may have typos. Check you third line. What is de?

[Unity+]

Give an explicit form each of the terms:

 

a(1) = 1

a(2) = 3

a(n+1) = [a(n) + a(n+1)]/2

 

I just started with this topic. How can I calculate what a(n) is?

 

Should I replace a(n) by 1 and a(n+1) by 3 and divide the result by 2? Would that give me de?

I am assuming that this is a recursive function.

 

a(1) would be a(n)

a(3) would be a(n+1)

 

Therefore, a(3) = (1+3)/2 = 2

 

If you notice, one of them is already a(2).

 

a(1) = 1

a(2) = 3

a(3) = (1 + 3)/2 = 2

a(2) = 3

a(3) = (1+3)/2 = 2

a(2) = 3

a(3) = 2

etc. etc

 

That is how I interpreted the problem.

[Klaynos]

I am assuming that this is a recursive function.

 

a(1) would be a(n)

a(3) would be a(n+1)

 

Therefore, a(3) = (1+3)/2 = 2

 

If you notice, one of them is already a(2).

 

a(1) = 1

a(2) = 3

a(3) = (1 + 3)/2 = 2

a(2) = 3

a(3) = (1+3)/2 = 2

a(2) = 3

a(3) = 2

etc. etc

 

That is how I interpreted the problem.

Your substitution didn't work.

 

You have a(n+1)=a(3) lhs, but then substitute in a(n+1)=a(2) in the rhs. You have defined n as both 1 and 2.

[Rapper]

Oh, I made a mistake, sorry...
It's

 

a(n+2) = [a(n) + a(n+1)]/2

How can I calculate a(5)?

[Klaynos]

Let n=1 and substitute into your equation.

 

Then let n=3 (you'll probably find you need n=2 too).

[imatfaal]

Try to do what Klaynos has suggested - and post your answers/workings; hopefully as soon as you do you will realise you are correct otherwise someone will set you back on the right course. But to help we need to see what you are getting to.

 

 

O/T funnily enough this works out as the same question (plus a constant of 2) to Function's question of a few weeks ago