a. Notice that
1/(1*2) = 1/2 = 1 - 1/2
1/(2*3) = 1/6 = 1/2 - 1/3
1/(3*4) = 1/12 = 1/3 - 1/4
and so on, which suggests the n-th term of the sum can be written as
[tex]\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}[/tex]
Then the sum itself is telescoping:
[tex]\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\dfrac1{3\cdot4}+\cdots+\dfrac1{n(n+1)}[/tex]
[tex]=\left(1-\dfrac12\right)+\left(\dfrac12-\dfrac13\right)+\left(\dfrac13-\dfrac14\right)+\cdots+\left(\dfrac1n-\dfrac1{n+1}\right)[/tex]
[tex]=1-\dfrac1{n+1}[/tex]
b. The proof is trivial:
[tex]\dfrac1n-\dfrac1{n+1}=\dfrac{n+1}{n(n+1)}-\dfrac n{n(n+1)}=\dfrac{n+1-n}{n(n+1)}=\dfrac n{n+1}[/tex]
so the formula found in (a) is correct.
The formula for series [tex]\mathbf{\frac{1}{1 \times 2},\frac{1}{2 \times 3}+...+\frac{1}{n(n+1)}}[/tex] is the sum of the series.
The expression for small n values is [tex]\mathbf{S_n = \frac{n}{n+1}}[/tex]
The series is given as:
[tex]\mathbf{\frac{1}{1 \times 2},\frac{1}{2 \times 3}+...+\frac{1}{n(n+1)}}[/tex]
(a) The expression for small n values
The series can be split as follows:
[tex]\mathbf{T_1 = \frac{1}{1 \times 2} = \frac{1}{1} - \frac{1}{2}}[/tex]
[tex]\mathbf{T_2 = \frac{1}{2 \times 3} = \frac{1}{2} - \frac{1}{3}}[/tex]
Express 3 as 2 + 1
[tex]\mathbf{T_2 = \frac{1}{2 \times (2 + 1)} = \frac{1}{2} - \frac{1}{2 + 1}}[/tex]
Express 2 as n
[tex]\mathbf{T_n = \frac{1}{n \times (n+1)} = \frac{1}{n} - \frac{1}{n + 1}}[/tex]
So, the sum is represented as:
[tex]\mathbf{S_n = (\frac{1}{1} - \frac{1}{2}) + (\frac 12 - \frac 13) +..........+(\frac 1n - \frac{1}{n+1})}[/tex]
Simplify
[tex]\mathbf{S_n = (\frac{1}{1} - \frac{1}{n+1})}[/tex]
[tex]\mathbf{S_n = 1 - \frac{1}{n+1}}[/tex]
Take LCM
[tex]\mathbf{S_n = \frac{n +1 - 1}{n+1}}[/tex]
[tex]\mathbf{S_n = \frac{n}{n+1}}[/tex]
Hence, the expression for small n values is [tex]\mathbf{S_n = \frac{n}{n+1}}[/tex]
(b) Prove (a)
Recall that:
[tex]\mathbf{T_n = \frac{1}{n} - \frac{1}{n + 1}}[/tex]
Take LCM
[tex]\mathbf{T_n = \frac{n + 1 - n}{n + 1}}[/tex]
Rewrite as:
[tex]\mathbf{T_n = \frac{n - n+ 1 }{n + 1}}[/tex]
Evaluate like terms
[tex]\mathbf{T_n = \frac{ 1 }{n + 1}}[/tex]
The expression n times is:
[tex]\mathbf{S_n = \frac{ 1 }{n + 1 }\times n}[/tex]
[tex]\mathbf{S_n = \frac{ n }{n + 1 }}[/tex]
Hence, the expression in (a) has been proved.
Read more about series at:
https://brainly.com/question/12429779
