Printable View
Hey all!
I needed some help in how to get the nth term (Un) for some series, that are:
1/1 + 3/2 + 5/2^2 + 7/2^3 + ...
1/2 + 2/3 + 3/4 + 4/5 + ...
1/5 + 2/6 + 2^2/7 + 2^3/8 + 2^4/9 + ...
I know the nth term (Un) for each but I wanted to know how you get them.
Thank you
Basicaly by inspection, and making some assumptions.Quote:
Originally Posted by dadon
Look at the first:
1/1 + 3/2 + 5/2^2 + 7/2^3 + ...
The numerators are consecutive odd numbers, which we may write as 2k-1
for k=1, 2, .. The denominators are consecutive powers of 2 which we may
write 2^(k-1) k=1, 2, .. (the -1 in the exponent is required here as I am
taking the first term to correspond to k=1 for both the numerator and
denominator), so the k-th term is (2k-1)/2^{k-1}.
Now look at the second:
1/2 + 2/3 + 3/4 + 4/5 + ...
The numerators are consecutive integers which we may write k, k=1, 2, ..
The denominators are also consecutive integers, but starting from 2, so
may be written (k+1), k=1, 2, .. Hence the general term is k/(k+1).
RonL
Hello, dadon!
These have pretty obvious patterns . . .
Quote:
1/1 + 3/2 + 5/2² + 7/2³ + ...
The numerators are: 1, 3, 5, 7, ...
. . The general numerator is: 2n - 1
The denominators are: 2^0, 2^1, 2^2, 2^3, ...
. . The general denominator is: 2^{n-1}
The general term is: .an .= .(2n - 1) / 2^{n-1}
Quote:
1/2 + 2/3 + 3/4 + 4/5 + ...
The numerators are: 1, 2, 3, 4, ...
. . The general numerator is: n
The denominators are: 2, 3, 4, 5, ...
. . The general denominator is: n + 1
The general term is: .an .= .n / (n + 1)
Quote:
1/5 + 2/6 + 2²/7 + 2³/8 + ...
The numerators are: 2^0, 2^1, 2^2, 2^3, ...
. . The general numerator is: 2^{n-1}
The denominators are 5, 6, 7, 8, ...
. . The general denominator is: n + 4
The general term is: .an .= .2^{n-1} / (n + 4)
Thank you!
Yeah does seem quite straight forward now. I was getting confused on the 2^(n-1) and was forgetting the 2^0, 2^1 terms.
Thanks again
Kind regards,
Dadon