Consider the following infinite series. a.Find the first four partial sums S_1,S_2,S_3, and S_4 of the series. b.Find a formula for the nth partial su

vazelinahS

vazelinahS

Answered question

2020-11-08

Consider the following infinite series.
a.Find the first four partial sums S1,S2,S3, and S4 of the series.
b.Find a formula for the nth partial sum Sn of the indinite series.Use this formula to find the next four partial sums S5,S6,S7 and S8 of the infinite series.
c.Make a conjecture for the value of the series.
k=12(2k1)(2k+1)

Answer & Explanation

Elberte

Elberte

Skilled2020-11-09Added 95 answers

Step 1.
We have to consider the infinite series:
k=12(2k1)(2k+1)
(a) To find the first four partial sums S1,S2,S3,S4 of the series.
b.Find a formula or the nth partial sum Sn of the infinite series.Use this formula to find the next four partial sums S5,S6,S7 and S8 of the infinite series.
c. Make a conjecture for the value of the series
Step 2
k=12(2k1)(2k+1)
(a) To find the first four partial sums S1,S2,S3,S4 of the series.
S1=a1=21(3)=23
S2=S1+a2=23+23.5=23+215=45
S3=S2+a3=45+25.7=45+235=67
S4=S3+a4=67+27.9=67+263=89
Step 3
(b) From the partial sums S1,S2,S3,S4 we see that the numerator is 2n and the denominator is 2n+1
So, we can write
Sn=2n2n+1
Now we find S5,S6,S7,S8
Sn=2n2n+1
S5=1011
S6=1213
S7=1415
S8=1617
(c) To make a conjecture of the value of the series.
k=12(2k1)(2k+1)
We can also write k=12(2k1)(2k+1) as k=12(4k21)
k=12(4k21)=k=1(12(k+12)+12(k12))
=(113)+(1315)+(1517)+(1719)+...
On cancelling out the terms we only left wiyh 1, So
k=12(2k1)(2k+1)=1
Therefore, the value of the series is 1.

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