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[recurrence relation] My solution differs from the book. Is there is error with my solution or is the book wrong?

13 viewed last edited 1 year ago
Hanna Owens
1
Problem: a_{n} = 2a_{n-1} + 5 and a_{0} = 3 I noticed this follows the pattern: 2^n \times a_{n} + 2^n-1 (5) + 2^n-2 (5) + ... + 2^1 (5) which turns into = 2^n (3) + 2^n-1 (5) + 2^n-2 (5) + ... + 2^1 (5) and this is where my solution begins to differ from the book. If I look at: 2^n-1 + 2^n-2 + ... + 2^1 This appears to be [∑ k=1, n-1 2^k ] So here is my question, what does this sum equal? I believe it is 2^n - 2 So carrying on with the computations I get 2^n (3) + 5 * 2^n - 2 = 3(2^n ) + 5 * (2^n - 2) =3(2^n ) + 5 * (2^n ) - 10 = 2^n+3 - 10 the book instead says the solution is... = 2^n (3) + 5 * (2^n - 1) = 3(2^n ) + 5 * (2^n ) - 5 = 2^n+3 - 5 Which is right? If I am wrong, where did I go wrong and how do I fix it?
laxman
0
2^{n-1} + 2^{n-2} + 2^{n-3}............ + 2^3 + 2^2 + 2^1 Actually this is in a geometric progression \text{a = 2 , the common ration r} = \frac{2^{n-2}}{2^{n-1}} = 2 The sum of the n-1 terms in geometric progression S_n= a* (\frac{1-r^n}{1-r}) So here: a = 2, r = 2, n = n-1 S_n = a* (\frac{1 - r^{n-1}}{1 - r}) S_n = 2* (\frac{1 - 2^{n-1}}{1 - 2}) S_n = 2* (\frac{1 - 2^{n-1}}{-1}) S_n = 2* ( 2^{n-1} - 1) S_n = 2^n - 2