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=1∞2(2k−1)(2k+1)

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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=1∞2(2k−1)(2k+1)

Elberte · Accepted Answer

Step 1. We have to consider the infinite series: ∑k=1∞2(2k−1)(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=1∞2(2k−1)(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=1∞2(2k−1)(2k+1) We can also write ∑k=1∞2(2k−1)(2k+1) as ∑k=1∞2(4k2−1) ∑k=1∞2(4k2−1)=∑k=1∞(−12(k+12)+12(k−12)) =(1−13)+(13−15)+(15−17)+(17−19)+... On cancelling out the terms we only left wiyh 1, So ∑k=1∞2(2k−1)(2k+1)=1 Therefore, the value of the series is 1.

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

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