Write as a series and express as a rational number:
1. 5.36363636....
2. 0.0123123....
Use this series and find S1,S2,S3,S4,Sn, and lim Sn.
1/1*3 + 1/3*5 + 1/5*7+...+ 1/(2n-1)(2n+1)
It works easier to have one type problem per question, as you will see here.
On the first, try something like multiping by a power of ten that will align digits. For instance, in the first, multiply by 100
536.36363636 . Subtract the original number
536.3636-5.363636= 531
Now divide by 99 (think out why)
531/99 is your fraction
I would split it this way:
5.36363636....
= 5 + (.36 + .0036 + .000036 + ...}
so for the bracket part, a=.36, r = .01
remember S∞ = a/(1-r)
= .36/(1-.01) = .36/.99 = 36/99 = 4/11
then 5.36363636.... = 5 4/11 or 59/11
do the rest the same way
1. 5.36363636.... = 5 + 0.36(1 + 10^-2 + 10^-4 + ..)
= 5 + 0.36[1/(1-.01)]
= 5 + 0.36 * 100/99
= 5 + 36/99
2. Do it the same way
3. If you are dealing with sums, the first term is
S1 = 1/3
and the second term is
S2 = 1/3 + 1/15 = 6/15
Do the others and see what Sn and the limit are
for your last question...
1/1*3 + 1/3*5 + 1/5*7+...+ 1/(2n-1)(2n+1)
S1 = 1/3
S2 = 1/3 + 1/15 = 6/15 = 2/5
S3 = 2/5 + 1/35 = 15/35 = 3/7
do you see a pattern?
so what is Sn ?
An interesting question now would be,
Prove that your answer to the above is correct by induction.
To express the given series as a rational number, we need to find the pattern in the repeating decimals and convert it into fractional form.
1. For the series 5.36363636..., we observe that the decimal part repeats itself as ".3636..." Let's call this repeating part "x". To get rid of the repeating part, we can multiply the entire number by a power of 10 that eliminates the repetition. In this case, we will multiply by 100 to remove the repeating decimal.
Let y = the given number 5.36363636...
Multiply both sides by 100:
100y = 536.363636...
Subtract the original equation from the multiplied equation to eliminate the repeating part:
100y - y = 536.363636... - 5.363636...
99y = 531
Divide both sides by 99 to get y alone:
y = 531/99
Therefore, the rational number form of 5.36363636... is 531/99.
2. Similarly, for the series 0.0123123..., we observe that the decimal part repeats itself as ".0123..." Let's call this repeating part "x". To eliminate the repeating part, we will multiply the entire number by a power of 10 that removes the repetition. In this case, we will multiply by 10000 since there are four digits in the repeating part.
Let y = the given number 0.0123123...
Multiply both sides by 10000:
10000y = 123.123123...
Subtract the original equation from the multiplied equation to eliminate the repeating part:
10000y - y = 123.123123... - 0.0123123...
9999y = 123
Divide both sides by 9999 to get y alone:
y = 123/9999
Therefore, the rational number form of 0.0123123... is 123/9999.
Next, let's calculate S1, S2, S3, S4, Sn, and lim Sn for the given series:
The given series is:
S = 1/1*3 + 1/3*5 + 1/5*7 + ... + 1/(2n-1)(2n+1)
1. Calculate S1:
S1 = 1/1*3 = 1/3
2. Calculate S2:
S2 = 1/1*3 + 1/3*5 = 1/3 + 1/15 = 6/15 = 2/5
3. Calculate S3:
S3 = 1/1*3 + 1/3*5 + 1/5*7 = 1/3 + 1/15 + 1/35 = 2/5 + 1/35 = 14/35 = 2/5
4. Calculate S4:
S4 = 1/1*3 + 1/3*5 + 1/5*7 + 1/7*9 = 1/3 + 1/15 + 1/35 + 1/63 = 2/5 + 1/35 + 1/63 = 42/105 + 3/105 = 45/105 = 3/7
5. Calculate Sn:
Sn = 1/1*3 + 1/3*5 + 1/5*7 + ... + 1/(2n-1)(2n+1)
= (2/5) for n ≥ 3, since the series repeats from S3 onwards.
6. Calculate lim Sn:
As n approaches infinity, Sn tends to a fixed value since the series starts repeating. In this case, lim Sn = 2/5.