Terms of a Sequence

[Jason76]

Write out 10 terms of this sequence.

\(\displaystyle a_{1} = 1\), \(\displaystyle a_{n+1} = a_{n} + (\dfrac{1}{2^{n}})\)

 

[HallsofIvy]

This is just and exercise in "Doing what you are told!"

\(\displaystyle a_1= 1\), \(\displaystyle a_2= a_1+ \frac{1}{2^1}= 1+ \frac{1}{2}= \frac{3}{2}\), \(\displaystyle a_3= a_2+ \frac{1}{2^2}= \frac{3}{2}+ \frac{1}{4}= \frac{7}{4}\)

Continue!

 

[stapel]

Write out 10 terms of this sequence.

\(\displaystyle a_{1} = 1\), \(\displaystyle a_{n+1} = a_{n} + (\dfrac{1}{2^{n}})\)

They gave you the value for the term when n = 1. For the next term, they told you to plug n = 1 into the right-hand side of the formula for all the terms other than the first term. So you plugged in the value for a₁, evaluated 1/(2¹), and simplified to find the value of a₂ = a₁₊₁. Then you repeated the process with n = 2 (for a₃ = a₂₊₁), and so forth.

Where are you stuck? Please be complete. Thank you!

 

[Ishuda]

Write out 10 terms of this sequence.

\(\displaystyle a_{1} = 1\), \(\displaystyle a_{n+1} = a_{n} + (\dfrac{1}{2^{n}})\)

There are (at least) two ways to go about the problem. One way is presented above. Another way is to note that, starting with the original equation, we have,
\(\displaystyle a_{n+1} = a_n + \frac{1}{2^n}\)
or, since
\(\displaystyle a_{n} = a_{n-1} + \frac{1}{2^{n-1}}\),
we have
\(\displaystyle a_{n+1} = a_{n-1} + \frac{1}{2^{n-1}} + \frac{1}{2^n}\)
Continuing in this fashion, we have
\(\displaystyle a_{n+1} = a_{n-2} + \frac{1}{2^{n-2}}+ \frac{1}{2^{n-1}} + \frac{1}{2^n}\)
...
\(\displaystyle a_{n+1} = a_1 + \frac{1}{2^1}+ \frac{1}{2^2} + ... + \frac{1}{2^{n-1}} + \frac{1}{2^n}\)
= \(\displaystyle \frac{1}{2^0} + \frac{1}{2^1}+ \frac{1}{2^2} + ... + \frac{1}{2^{n-1}} + \frac{1}{2^n}\)
or
\(\displaystyle a_{n+1} = \Sigma_{j=0}^{j=n} \space \frac{1}{2^j} = 2 - (\frac{1}{2})^{n}\)
where the last part of the equation is just the simplified formula for the sum of a geometric series with common ratio \(\displaystyle \frac{1}{2}\).