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Let $a_{1}, a_{2}, a_{3,} \ldots$ be a sequence defined by

$$a_{1}=1, a_{n}=2 a_{n-1}+1 ; n \geqslant 2$$

Show that $a_{n}=2^{n}-1$ for all positive integers $n>1$

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problem we want to use Ah, induction toe Prove that I am, which is found by the recruits of statement to a n minus one plus one can also be found by using the equation Heard expression to to the n minus one. So, uh, we'll do our base case first. The base cases for n equals 22 since it says, and greater than what? For an equals two on the left side, a two would be equal to two times a one plus one, which is two times a one is given to us. That's one plus one, which makes three. On the other side, we have to to the to minus Winston's, and that's equal to two that is four minus one, which is also equal to three. So since the left side and right side are equal, we know a one is true. So the inductive step assume that a K is true. So if a K is true, that means a K is defined by two times a K minus one the previous term plus one, which is equal to two times K minus one, then for a K plus one. Uh, that would be equal to two times the previous term, a k plus one a. K. We just found an expression for it. That's the same as two to the K minus one seven plus one. And this is exponents laws again. So two times serially kay would be to to the K plus one minus two plus one. So this is to to the K plus one minus one, and that satisfies the original expression for the case K plus one. Therefore, uh, a k plus one is true when a K is true. Therefore a to the end is true for all end in the positive integers where and is greater than one.

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