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Question: Consider the sequence defined as $a_1 = 2$ and $a_k = a_{k-1}+2k-1$ for all positive integer $ k \geq 2$ . Show that $a_n = 1+\sum(2i-1, i = 1 .. n)$ .
Hint: Start with $\sum(2i-1, i = 1 .. n)$ and use the recursive definition of the sequence.

Answer: I am unsure where to start on to show the proof. I can't find an example. I know the equation is $n^2+1$, I need to use the recursive definition. Any help would work. Thanks!

Donald Splutterwit
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Salan
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2 Answers2

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The proof is by induction on $n$, the base case being obvious. For the inductive step, for any $n\ge 1$, proceed as follows: \begin{align} a_{n+1} &=a_n+2(n+1)-1 \\&=1+\left(\sum_{k=1}^n2k-1\right)+2(n+1)-1 \\&=1+\left(\sum_{k=1}^{n+1}2k-1\right)\end{align} The last step follows from the inductive definition of $\sum_{k=1}^n$; by definition, $\sum_{k=1}^{n+1}b_k=\left(\sum_{k=1}^n b_k\right)+b_{n+1}$ holds for all $n\ge 1$. Let $b_k=2k-1$.

Mike Earnest
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you may proceed as follows:

\begin{array} aa_n & = a_{n-1}&+ 2n&- 1\\ a_{n-1} & = a_{n-2}&+2(n-1)&-1 \\ \ldots&=\ldots&+\ldots&-1\\ a_2 &=a_1&+2(2)&-1 \end{array}

Add all rows and see what you get. Further simplification will will you $$a_n=2\Big(\frac{n(n+1)}{2}\Big)-n+1=n^2+1$$

Mittens
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