Question

Find the ${{n}^{th}}$ term of the series $2+5+12+31+86+......$

Answer 1

We first try to find the ${{n}^{th}}$ term of the series using the subtraction form of the terms. In that case, we subtract shifting one term on the right side. We find the term form in general sequence and complete the process one more time to find the solution.

Complete step-by-step solution:
We have been given $2+5+12+31+86+......$. This is AGM progression.
We assume its ${{n}^{th}}$ term as ${{t}_{n}}$. So, $2+5+12+31+86+......+{{t}_{n}}$
We assume the sum as $S=2+5+12+31+86+......+{{t}_{n}}$.
We apply one particular trick to subtract S from S in a particular way taking crosswise subtraction as
$\begin{aligned} & S=2+5+12+31+86+......+{{t}_{n}} \\\ & \underline{S= 0+2+5+12+31+86+......+{{t}_{n}}} \\\ & 0=2+\left( 5-2 \right)+\left( 12-5 \right)+\left( 31-12 \right)+\left( 86-31 \right)+......+\left( {{t}_{n}}-{{t}_{n-1}} \right)-{{t}_{n}} \\\ \end{aligned}$
The series has no terms. The simplified form is $2+3+7+19+55+......+\left( {{t}_{n}}-{{t}_{n-1}} \right)-{{t}_{n}}=0$.
So, ${{t}_{n}}=2+3+7+19+55+......+\left( {{t}_{n}}-{{t}_{n-1}} \right)$ and it has n terms.
We perform the same process one more time on the series of $3+7+19+55+......+\left( {{t}_{n}}-{{t}_{n-1}} \right)$.
This series has $n-1$ terms where we assume its ${{n}^{th}}$ term as ${{T}_{n}}$ and $\left( {{t}_{n}}-{{t}_{n-1}} \right)={{T}_{n-1}}$.
We assume the sum as ${{S}^{'}}=3+7+19+55+......+{{T}_{n-1}}$.
$\begin{aligned} & {{S}^{'}}=3+7+19+55+......+{{T}_{n-1}} \\\ & \underline{{{S}^{'}}=0+3+7+19+55+......+{{T}_{n-1}}} \\\ & 0=3+\left( 7-3 \right)+\left( 19-7 \right)+\left( 55-19 \right)+......+\left( {{T}_{n-1}}-{{T}_{n-2}} \right)-{{T}_{n-1}} \\\ \end{aligned}$
The series has $n-1$ terms. The simplified form is $3+4+12+36+......+\left( {{T}_{n-1}}-{{T}_{n-2}} \right)-{{T}_{n-1}}=0$
So, ${{T}_{n-1}}=3+4+12+36+......+\left( {{T}_{n-1}}-{{T}_{n-2}} \right)$ and it has $n-1$ terms.
We exclude the term 3 at the start and get a GP series. The first term of the series is 4 and the common ratio is $\dfrac{12}{4}=3$. The GP series is for $n-2$ terms.
The value of $\left| r \right|>1$ for which the sum of the first n terms of an G.P. is ${{S}_{n}}={{a}_{1}}\dfrac{{{r}^{n}}-1}{r-1}$.
The sum is ${{T}_{n-1}}=3+4\times \dfrac{{{3}^{n-2}}-1}{3-1}=3+2\left( {{3}^{n-2}}-1 \right)=1+2\times {{3}^{n-2}}$.
So, ${{S}^{'}}=3+7+19+55+......+{{T}_{n-1}}=3+7+19+55+......+\left( 1+2\times {{3}^{n-2}} \right)$.
We can write as ${{S}^{'}}=3+7+19+55+......+\left( 1+2\times {{3}^{n-2}} \right)=\sum\limits_{r=2}^{n}{\left( 1+2\times {{3}^{r-2}} \right)}$.
We get ${{S}^{'}}=\sum\limits_{r=2}^{n}{\left( 1+2\times {{3}^{r-2}} \right)}=\left( n-1 \right)+2\times \dfrac{{{3}^{n-1}}-1}{3-1}={{3}^{n-1}}+n-2$.
So, ${{t}_{n}}=2+{{S}^{'}}=2+{{3}^{n-1}}+n-2={{3}^{n-1}}+n$.
Therefore, the ${{n}^{th}}$ term of the series $2+5+12+31+86+......$ is ${{t}_{n}}={{3}^{n-1}}+n$.

Note: This special form of subtraction is done to get the ${{t}_{n}}$ in negative form with the number of terms in that subtraction being equal to the number of terms in the main series. The sequence is an increasing sequence where the common ratio is a positive number. The common difference will never be calculated according to the difference of greater number from the lesser number.