General term for increasing AP's

Question

Can some give an easy general method to find general term of sequences whose difference is in AP? Example: 1,4,8,13,19....

The difference is 3,4,5... which is in AP.

Through vigorous testing and analysis you get the general term as $ (n^2 + 3n - 2)/2$ but how to derive it mathematically

Answer 1

Let us first closely examine the sequence: $1,4,8,13,19,\dots$ \begin{align*} u_1&=1\\ u_2&=1+3=4\\ u_3&=1+(3+4)=8\\ u_4&=1+(3+4+5)=13\\ \vdots \end{align*} Notice that essentially to get the term $u_n$ in our sequence, we need to add the sum of the first $n-1$ terms in the difference sequence ($3,4,5,6,\dots$) to the first term $u_1=1$. So, to find the general expression for $u_n$, we first need find the general expression for the sum of the first $n-1$ terms in the difference sequence.

We have $a_1=3$ and $d=1$, meaning $a_n=a_1+d(n-1)=3+1\cdot(n-1)=n+2$. So, if $$S_{n}=\frac{a_1+a_n}{2}\cdot n=\frac{n(3+n+2)}{2}=\frac{n(n+5)}{2}$$ Then $$S_{n-1}=\frac{(n-1)(n-1+5)}{2}=\frac{(n-1)(n+4)}{2}$$ Now, we have the first term of the original sequence $u_1=1$, we know the difference expression that is added to the first term to get the $n$th term, so let us put it all together in a general expression for $u_n$! \begin{align*} u_n&=u_1+S_{n-1}\\ &=1+\frac{(n-1)(n+4)}{2}\\ &=\frac{2+n^2+4n-n-4}{2}\\ &=\frac{n^2+3n-2}{2} \end{align*} This is exactly what you have obtained by testing.

Returning to your question, the same steps can be taken for a general sequence, if its difference is in arithmetic progression. Namely, you find the first term of the original sequence, then the general expression for the sum of the first $n-1$ terms for the difference sequence, and put it together. If you want an exercise, try to work out the expression for the term $u_n$ for a general sequence.

Hope this helps and please feel free to ask any further questions, if anything is a bit unclear! :)

Answer 2

Yes dude you got the right question. There is a method to obtain general term directly for your question.This method can even extended when difference of differences are in AP or even further. To find $t_1 + t_2 + t_3 + \cdots +t_n$

Let $S_n = t_1+t_2+t_3+ \cdots + t_n$

Then find, $ \Delta t_1, \Delta t_2 ,\Delta t_3 $ [first order differences] $\Delta^2 t_1 ,\Delta^2 t_2 ,\Delta^2 t_3$[2nd order differences] ...Continue the process until you get all the terms 0. Note : here $\Delta t_1 = t_2-t_1, \Delta t_2 = t_3-t_2 \space etc.$ and $\Delta^2 t_1 = \Delta t_2 -\Delta t_1, \Delta^2 t_2 = \Delta t_3 - \Delta t_2\space etc.$ Similarly you can find higher order differences.After obtaining all the required order of differences you can write $t_n $ as $t_n = \binom{n-1}{0}t_1 + \binom{n-1}{1}\Delta t_1 + \binom{n-1}{2}\Delta^2 t_1......$ and $S_n = \binom{n}{1}t_1+\binom{n}{2} \Delta t_1 + \binom{n}{3} \Delta^2 t_1....$ You can apply this process for your question and get the same as you got.Note use this process only when you see that the difference of differences of even further are in AP.Not any other sequence.

Answer 3

Let $a_1$, $a_2$, $\dots$, be a sequence such that the sequence $a_2-a_1$, $a_3-a_2$, $\dots$, forms an AP with first term $b$ and common difference $d$. Then, $$a_n-a_1=(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+\dots+(a_2-a_1)$$ So, $a_n-a_1$ is the sum of first $n-1$ terms of this new AP. I assume you know what the sum of $n$ terms of an AP is. So, $$a_n=a_1+\frac{n-1}{2}(2b+(n-2)d)$$ Plugging in $a_1$, $b$ and $d$, you can find the general term.

In your case, $a_1=1$, $b=3$ and $d=1$ gives $$a_n=1+\frac{n-1}{2}(2\times3+(n-2))$$ so $$a_n=1+\frac12(n-1)(n+4)$$ which you can see is precisely the same as what you have got.

Hope this helps. :)

Answer 4

Since the differences are in AP, it means that the sequence is described by a quadratic function. This is because the first difference of a linear function is constant, and the second difference of a quadratic function is constant.

Let's denote the general term of the sequence term of the sequence by $a_n$ and the general term of the difference sequence (i.e. the $nth$ difference) by $d_n$. Given that $d_n$ is an AP, we can write it as:

$$d_n = d_1 + (n - 1)d$$

where $d_1$ is the first difference (which is 3 in your example) and $d$ is the common difference of the differences (which is 1 in your example).

Now, since $a_n - a_{n-1} = d_n$, we can express $a_n$ recursively as:

$$ a_n = a_{n-1} + d_n $$

We can start building the expression for $a_n$ by summing up the differences:

$$ a_n = a_1 + \sum_{i=1}^{n-1} d_i $$

We know that $$d_n = d_1 + (n - 1)d$$ or rather $$d_i = d_1 + (i - 1)d$$

Hence

$$ a_n = a_1 + \sum_{i=1}^{n-1} [d_1 + (i - 1)d] $$

We know that $a_1 = 1$, $d_1 = 3$ and $d = 1$, so

$$ a_n = 1 + \sum_{i=1}^{n-1} [3 + (i - 1)] $$

$$ a_n = 1 + \sum_{i=1}^{n-1} 3 + \sum_{i=1}^{n-1} (i - 1) $$

$$ = 1 + \sum_{i=1}^{n-1} 3 + \sum_{i=1}^{n-1} i - \sum_{i=1}^{n-1} 1 $$

$$ = 1 + \sum_{i=1}^{n-1} 3 - \sum_{i=1}^{n-1} 1 + \sum_{i=1}^{n-1} i $$

$$ a_n = 1 + \sum_{i=1}^{n-1} 2 + \sum_{i=1}^{n-1} i $$

by the distributive property of summation.

$\sum_{i=1}^{n-1}i$ is the sum of the first $n-1$ natural numbers, where $u_1 = 1$ and $d = 1$.

Substituting in the summation of an arithemtic sequence formula $$S_n = \frac{n}{2} \cdot (2u_1 + (n - 1)d)$$

$$\sum_{i=1}^{n-1}i = S_{n-1}$$ $$ = \frac{n-1}{2} \cdot (2 + ((n-1) - 1))$$ $$ = \frac{n-1}{2} \cdot (2 + (n-1-1))$$ $$ = \frac{n-1}{2} \cdot (2+n-2)$$ $$ = \frac{n-1}{2} \cdot n$$ $$ \sum_{i=1}^{n-1}i= \frac{n(n-1)}{2}$$

Remember that $$ a_n = 1 + \sum_{i=1}^{n-1} 2 + \sum_{i=1}^{n-1} i $$

We know that $$\sum_{i=1}^{n-1} 2 = 2(n-1)$$ as it is just $2$ summed $n-1$ times.

We have also found that $$ \sum_{i=1}^{n-1}i= \frac{n(n-1)}{2}$$

Substituting both of these into our equation for $a_n$:

$$ a_n= 1 + 2(n-1) + \frac{n(n-1)}{2} $$ $$ = \frac{2}{2} + \frac{4(n-1)}{2} + \frac{n(n-1)}{2} $$ $$= \frac{2 + 4(n-1) + n(n-1)}{2} $$ $$= \frac{2 + 4n-4 + n^2 -n}{2} $$ $$= \frac{-2 + 3n+n^2}{2} $$ $$= \frac{n^2 + 3n - 2}{2}$$ $$ a_n = \frac{n^2 + 3n - 2}{2}$$