Find 10 partial sums of the series. (round your answers to five decimal places.) \(\sum _{n=1}^{\infty }\) 15(-4)n?
Solution:
\(\sum _{n=1}^{\infty }\) 15(-4)n
Now, we have to find 10 partial sums of the series,
S1 = \(\sum _{n=1}^{1}\)15(-4)n = 15(-4) = -60
S2 = \(\sum _{n=1}^{2}\)15(-4)n = \(\sum _{n=1}^{1}\)15(-4)n + 15(-4)2 = - 60 + 15(16) = -60 + 240 = 180
S3 = \(\sum _{n=1}^{3}\)15(-4)n = \(\sum _{n=1}^{2}\)15(-4)n + 15(-4)3 = 180 + 15(-64) = 180 - 960 = -780
S4 = \(\sum _{n=1}^{4}\)15(-4)n = \(\sum _{n=1}^{3}\)15(-4)n + 15(-4)4 = -780 + 15(256) = -780 - 3840 = 3060
S5 = \(\sum _{n=1}^{5}\)15(-4)n = \(\sum _{n=1}^{4}\)15(-4)n + 15(-4)5 = 3,060 + 15(-1024) = 3060 - 15,360 = -12,300
S6 = \(\sum _{n=1}^{6}\)15(-4)n = \(\sum _{n=1}^{5}\)15(-4)n + 15(-4)6 = -12,300 + 15(4096) = -12,300 + 61,440 = 49,140
S7 = \(\sum _{n=1}^{7}\)15(-4)n = \(\sum _{n=1}^{6}\)15(-4)n + 15(-4)7 = 49,140 + 15(-16,384) = 49,140 - 2,45,760 = -1,96,620
S8 = \(\sum _{n=1}^{8}\)15(-4)n = \(\sum _{n=1}^{7}\)15(-4)n + 15(-4)8 = -1,96,620 + 15(65,536) = -1,96,620 + 9,83,040 = 7,86,420
S9 = \(\sum _{n=1}^{9}\)15(-4)n = \(\sum _{n=1}^{8}\)15(-4)n + 15(-4)9 = 7,86,420 + 15(-2,62,144) = 7,86,420 - 39,32,1600 = -31,45,740
S10 = \(\sum _{n=1}^{10}\)15(-4)n = \(\sum _{n=1}^{9}\)15(-4)n + 15(-4)10 = -31,45,740 + 15(10,48,576) = -31,45,740 + 1,57,28,640 = 1,25,82,900
Therefore, the 10 partial sums of the series are - 60, 180, -780, 3060, -12,300, 49,140, -1,96,620, 7,86,420, -31,45,740, 1,25,82,900.
Find 10 partial sums of the series. (round your answers to five decimal places.) \(\sum _{n=1}^{\infty }\) 15(-4)n?
Summary:
The 10 partial sums of the series are - 60, 180, -780, 3060, -12,300, 49,140, -1,96,620, 7,86,420, -31,45,740, 1,25,82,900.
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