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So i was doing this question on my workbook. I did the first question and i was correct. but on the second question, same logic, same solving way, but i was wrong. here is question 1, finding a closed form for $$a_1=3\qquad a_{n+1}=a_n+2,n\ge2$$ Thus I have $$a_{n+1}=a_n+2,d=2$$ $$a_n=a_{n-1}+2$$ $$a_n=a_1+(n-1)*2$$ $$a_n=a_1+2n-2$$ $$a_1=3$$ $$a_n=3+2n-2$$ $$a_n=2n+1,n\ge1$$ the last line was the answer, and i did it right.but the same logic couldn't be applied here $$a_1=1$$ $$a_n=a_{n-1}+2n-1,n\ge2$$ heres my solution $$d=2n-1$$ $$a_n=a_1+(n-1)*(2n-1)$$ $$a_n=a_1+2n^2-3n+1$$ $$a_1=1$$ $$a_n=2n^2-3n+2$$ Then i couldn't simplify it futher. But the answer says else-wise $$a_n=n^2$$

Parcly Taxel
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doodler doodle
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  • The second example is not an arithmetic progression, hence you can not just use the formulas attached to arithmetic progressions. – lulu Jun 29 '19 at 14:46
  • The second case isn't an arithmetic progression, so the step $a_n=a_1+(n-1)*(2n-1)$ is incorrect. – saulspatz Jun 29 '19 at 14:47
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    +1 Welcome to MSE. My compliments on a well-formulated question. You stated the problem clearly and completely, and showed all your work. Keep it up! – saulspatz Jun 29 '19 at 14:51

3 Answers3

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Hint:

Remember that an arithmetic progression is a series where two consecutive terms have the same common difference.

Also, note that $a_n,a_{n+1}$ are two consecutive terms and $a_{n-1},a_n$ are two consecutive terms.

So, by definition, $a_n-a_{n-1}$ and $a_{n+1}-a_n$ should be constant common difference.

However, for the second question, the common difference is dependent on $n$ and thus not constant. So the second one is not an AP.

Hoewvwer an alternate approach to solve the second problem exists.

Using the fact $a_n-a_{n-1}=2n-1$ we get,

$$a_2-a_1=3$$ $$a_3-a_2=5$$ $$a_4-a_3=7$$ $$.$$ $$.$$ $$a_n-a_{n-1}=2n-1$$

Adding the two columns we have $$a_n-a_1=3+5+7+...+(2n-1)$$ $$\implies a_n=1+3+5+7+...+(2n-1)\text{ [since a_1=1]}$$

So now the right side is an AP.

Add it and voila!

Soham
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One way is to note, that that because n is variable, differences are in arithmetic progression in this case. This means, we can sum up all the differences as in an arithmetic progression. $n-1$ terms gives us ${(n-1)(3+(2n-1))\over 2}$ we can then add this to $a_1$ getting $1+(n-1)(n+1)= 1+n^2-1=n^2$

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You can't treat the second recurrence as an arithmetic progression since its step size isn't constant.

That $a_n=n^2$ there may perhaps be most easily seen visually:

...o... a_1
..oox.. a_2
.oooxx. a_3
ooooxxx a_4

and then you fold the triangle of x onto that of o, obtaining a square of size $n$. The formal proof may be achieved via induction.

Parcly Taxel
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