A Series with an increment in its difference .

Question

I came across this series , $$1⠀⠀_1⠀⠀2⠀⠀_2⠀⠀4⠀⠀_3⠀⠀7⠀⠀_4⠀⠀11 ⠀.⠀.⠀.⠀.⠀.⠀.$$ As the difference of the series is in $AP$ then , [$Notations:$ $A$ is first term of the series , $a$ is difference of first two terms $\&$ $d$ is the incement in the difference of the series .]

So , $T_n$ will be the sum of differences of the series plus first term of the series , $$T_n=A+\frac{(n-1)}{2}[2a+d(n-2)]$$

And , $S_n=\sum T_n$ , [ here , $m\in\mathbb N$ ]

\begin{equation} \label{eq1} \begin{split} S_n & =\sum_{m=1}^{n}\left(A+\frac{(m-1)}{2}[2a+d(m-2)]\right) \\ &⠀\\ & = An+a\sum_{m=1}^{n}(m-1)+\frac{d}{2}\sum_{m=1}^{n}(m-2)(m-1) \\ &⠀\\ &=An+a\left(\frac{n(n+1)}{2}-n\right)+\frac{d}{2}\left(\frac{n(n+1)(2n+1)}{6}-\frac{3n(n+1)}{2}+2n\right) &⠀\\ &⠀\\ &=An+\frac{an(n-1)}{2}+dn\left(\frac{(n+1)(2n+1)-9(n+1)+12}{12}\right)\\ &⠀\\ &=An+\frac{an(n-1)}{2}+\frac{dn(n-1)(n-2)}{6} &⠀\\ &=An+\frac{n(n-1)}{2}\left(a+\frac{d}{3}(n-2)\right) &= \end{split} \end{equation}

Now , we got : $$S_n=An+\frac{n(n-1)}{2}\left(a+\frac{d}{3}(n-2)\right)$$ My question is that when , $d(n-2)\not\equiv 0\mod3$ then , $S_n\notin\mathbb Z$ . Or basically $d$ and $(n-2)$ is not a multiple of $3$ then , How do we proceed to solution . For example , Find the $10^{th}$ of this series , $$1⠀2⠀5⠀10⠀17⠀.⠀.⠀.⠀.⠀.⠀T_{10}$$

Answer 1

We have a factor of $\frac{n(n-1)}{2}$ outside.

That is $$S_n = An +\frac{an(n-1)}{2}+\frac{dn(n-1)(n-2)}{6}$$

$n(n-1)(n-2)$ being the product of $3$ consecutive integers is divisible by $6$.

Answer 2

Substituting values in $S_n=An+\frac{n(n-1)}{2}\left(a+\frac{d}{3}(n-2)\right)$, we get

$S_n=n+\frac{n(n-1)}{2}\left(1+\frac{1}{3}(n-2)\right)$ which is further simplified as-

$S_n=n+\frac{n(n-1)}{2}\left(\frac{n+1}{3}\right)$

$S_n=n+\frac{n(n-1)(n+1)}{6}$

Now $\frac{n(n-1)(n+1)}{6}$ will always be divisible by $6$ as it is product of three consecutive numbers and out of these three numbers exactly one will be a multiple of 3 and at least one will be an even number. So this value will always simplify to a whole number for a natural number $n$. Therefore at last you will be left with with a natural number only.

$S_{10}=10+\frac{10(9)(11)}{6}=175$