Find the first three partial sums and the nth partial sums of the sequence
An=1/n+2-1/n+3
The sequence is represented by the formula An = 1/(n+2) - 1/(n+3).
To find the first three partial sums, we need to add up the terms of the sequence:
First partial sum:
A1 = 1/(1+2) - 1/(1+3) = 1/3 - 1/4 = 4/12 - 3/12 = 1/12
Second partial sum:
A1 + A2 = (1/3) + 1/(3+2) - 1/(3+3) = 1/3 + 1/5 - 1/6 = 10/30 + 6/30 - 5/30 = 11/30
Third partial sum:
A1 + A2 + A3 = (1/3) + (1/5) + 1/(5+2) - 1/(5+3) = 1/3 + 1/5 + 1/7 - 1/8 = 280/840 + 504/840 + 360/840 - 315/840 = 829/840
To find the nth partial sum, we can use a general formula:
nth partial sum = A1 + A2 + A3 + ... + An
With the given sequence An = 1/(n+2) - 1/(n+3), it is not possible to find a simple closed-form expression for the nth partial sum.
To find the partial sums of the sequence A_n = 1/(n+2) - 1/(n+3), we need to add up the terms of the sequence up to a certain number of terms.
The first three terms are:
A_1 = 1/(1+2) - 1/(1+3) = 1/3 - 1/4 = 1/12
A_2 = 1/(2+2) - 1/(2+3) = 1/4 - 1/5 = 1/20
A_3 = 1/(3+2) - 1/(3+3) = 1/5 - 1/6 = 1/30
The first three partial sums are:
S_1 = A_1 = 1/12
S_2 = A_1 + A_2 = 1/12 + 1/20 = 8/60 + 3/60 = 11/60
S_3 = A_1 + A_2 + A_3 = 1/12 + 1/20 + 1/30 = 10/60 + 3/60 + 2/60 = 15/60
To find the nth partial sum, we need to add up the terms of the sequence from A_1 to A_n. The nth partial sum is given by:
S_n = A_1 + A_2 + ... + A_n
For example, the fifth partial sum (S_5) would be the sum of the first five terms:
S_5 = A_1 + A_2 + A_3 + A_4 + A_5
To find a formula for the nth partial sum, let's simplify the sequence first. We can rewrite A_n as:
A_n = 1/(n+2) - 1/(n+3)
= (n+3)/(n+2)(n+3) - (n+2)/(n+2)(n+3)
= ((n+3) - (n+2))/(n+2)(n+3)
= 1/(n+2)(n+3)
Now we can find a formula for the nth partial sum:
S_n = A_1 + A_2 + ... + A_n
= 1/((1+2)(1+3)) + 1/((2+2)(2+3)) + ... + 1/((n+2)(n+3))
Therefore, the nth partial sum is given by the formula:
S_n = ∑ (1/((k+2)(k+3))), where k goes from 1 to n.