find the sum of the sequence and show work!! thanks!! (-1/2)^n n=5 to infinity S
ID: 3081765 • Letter: F
Question
find the sum of the sequence and show work!! thanks!! (-1/2)^n n=5 to infinityExplanation / Answer
Don't get hung up on the notation. Let me see if I can simplify it for you: Un = 0.8Un-1 + 4 This just means: New term = 0.8 · (Previous term) + 4 Since you need to know a previous term in order to get a new term, it's necessary to include an initial condition: U1 = 5 That means the first term = 5 Using the recurrence relation, Second term = 0.8 · (first term) + 4 .·. Second term = 0.8 · 5 + 4 = 4 + 4 = 8 The notation for the second term is U2 .·. U2 = 8 Now, you can work out by hand the answer to (b): U3 = 0.8·U2 + 4 = 6.4 + 4 = 10.4 U4 = 0.8·U3 + 4 = 8.32 + 4 = 12.32 U5 = 0.8·12.32 + 4 = 13.856 U6 = 0.8·13.856 + 4 = 15.0848 Therefore, the smallest n such that Un> 15 is n = 6 For evaluating the limit, you can see that the terms continue to grow, but at a slower and slower rate. Perhaps you can look at this analytically: U2 = 0.8·U1 + 4 U3 = 0.8·U2 + 4 = 0.8·(0.8·U1 + 4) + 4 = 0.8²·U1 + 0.8·4 + 4 U4 = 0.8·U3 + 4 = 0.8·( 0.8²·U1 + 0.8·4 + 4) + 4 = 0.8³ · U1 + 0.8²·4 + 0.8·4 + 4 = 0.8³ + 4·(0.8² + 0.8 + 1) U5 = 0.8·U4 + 4 = 0.8·(0.8³ · U1 + 0.8²·4 + 0.8·4 + 4) + 4 = 0.84 ·U1 + 4·(0.8³ + 0.8² + 0.8 + 1) From this, a pattern emerges: Un = 0.8^(n-1)·U1 + 4·(0.8^(n-2) + 0.8^(n-3) + ... + 0.8³ + 0.8² + 0.8 + 1) As n -->8, the first term goes to 0 (because 0.8^8 --->0) The second part, 4·(0.8^(n-2) + 0.8^(n-3) + ... + 0.8³ + 0.8² + 0.8 + 1) can be rewritten in its conventional form: 4 + 4·0.8 + 4·0.8² + 4· 0.8³ + ... + 4·0.8^(n-3) +4·0.8^(n-3), which is a geometric series. The first term is 4 and the constant ratio is 0.8 There is a nice little formula for the sum of an infinite geometric series with a first term, a, and a ratio, r. If |r| 8 S = 4 / (1 - 0.8) = 20Related Questions
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