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I have encountered an induction problem which I don't understand. What I don't understand is what it is asking me to prove. I don't want a solution. The problem is:

If $u_1=5$ and $u_{n+1}=2u_n-3(-1)^n$, then $u_n=3(2^n)+(-1)^n$ for all positive integers.

Am I supposed to prove $u_{n+1}=2u_n-3(-1)^n$ or $u_n=3(2^n)+(-1)^n$ is true for all positive integers?

Babelfish
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    You are supposed to prove that $u_n= 3(2^n)+(-1)^n$. $u_{n+1}=2u_n-3(-1)^n$ is a definition. – Matthew Towers Sep 07 '18 at 14:15
  • Induction is : [if $P(1)$ and "if $P(n)$, then $P(n+1)$, for every $n$", then "for every $n$, $P(n)$"]. – Mauro ALLEGRANZA Sep 07 '18 at 14:16
  • The expression for $u_n$ is the $P(n)$. First step: is it true that for $n=1$ its value is $5$ ? – Mauro ALLEGRANZA Sep 07 '18 at 14:17
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    $",\color{blue}{\style{font-family:inherit}{\text{If}};u_1=5 ;\style{font-family:inherit}{\text{and}} ;u_{n+1}=2u_n-3(-1)^n},, ;\color{red}{\style{font-family:inherit}{\text{ then}};u_n=3(2^n)+(-1)^n},",$ means you are given the blue part, and have to prove the red part. – dxiv Sep 07 '18 at 19:41

3 Answers3

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You are supposed to prove $u_n=3(2^n)+(-1)^n$.

$u_1=5$ and $u_{n+1}=2u_n-3(-1)^n$ are the conditions you are supposed to make use of.

Yuta
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$u_{n+1}=2u_n-3(-1)^n$ is your recurrence, specifying on how to obtain the $n+1$-st term $u_{n+1}$ from the $n$-th term $u_n$. Together with a start value, $u_1=5$, this stepwise determines the sequence completely.

What the problem is trying to establish, is to show that in general, for any $n$, you can obtain the $n$-th term directly(without evaluating all $u_k$'s with $k<n$ before) via the formula $u_n=3(2^n)+(-1)^n$. That this formula holds can be shown using induction.

blub
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You are given $u_{n+1}=2u_n-3(-1)^n$. This would allow you to compute the entire series. For example $$u_2=2u_1-3(-1)^1=2\cdot 5+3=13\\ u_3=2\cdot 13-3(-1)^2=23$$ and so on. You are supposed to prove the last statement. It claims $$u_1=3(2^1)+(-1)^1=5\\u_2=3(2^2)+(-1)^2=13\\u_3=3(2^3)+(-1)^3=23\\\vdots$$

user21820
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Ross Millikan
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