Question

Find the ${n^{th}}$ term of the series $1,3,8,16,27,41,...................$

Answer 1

For solving this problem first we write the given series and then in the next row below it we write the terms of the same series by shifting each term to the right by one place. Further we subtract the lower row from the upper row. The series so obtained is in A.P. After that we find the sum of this series up to n term. By this we get the $n^{th}$ term of the given series.

Complete step by step answer:
Given series $1,3,8,16,27,41,...................$
Since terms of given series are not in A.P. but the difference of two consecutive terms of the series is in A.P. then for getting the sum of given series we use Method of Difference.
In this method first we write the sum of the given series then in the next row below it we write the terms of the same series by shifting each term to the right by one place.
$S = 1 + 3 + 8 + 16 + 27 + 41 + ...................{T_n}$ ------------------(1)
$S = {\text{ }}1 + 3 + 8 + 16 + 27 + 41 + ...................{T_{n - 1}} + {T_n}$ ----------(1)
On subtracting equation $\left( 2 \right)$ from equation $\left( 1 \right)$,we get
$0 = 1 + 2 + 5 + 8 + 11 + 14 + ..................\left( {{T_n} - {T_{n - 1}}} \right) - {T_n}$
Now we shift ${T_n}$ on left side
${T_n} = 1 + (2 + 5 + 8 + 11 + 14 + ................upto \left( {n - 1} \right)terms)$
The series above inside the bracket is in A.P. so we find the sum of A.P. upto $(n - 1)$ terms
By using formula ${S_{n - 1}} = \dfrac{{n - 1}}{2}\left[ {2a + \left( {n - 2} \right)d} \right]$
Since from A.P. first term $a = 2$ and difference $d = 5 - 2 = 3$
${T_n} = 1 + \dfrac{{n - 1}}{2}\left[ {2 \times 2 + \left( {n - 2} \right) \times 3} \right]$

$$\Rightarrow {T_n} = 1 + \dfrac{{n - 1}}{2}\left[ {4 + 3n - 6} \right] \\\ \Rightarrow {T_n} = 1 + \dfrac{{(n - 1)(3n - 2)}}{2} \\\$$

On multiplying $(n - 1)$ and $(3n - 1)$
$\Rightarrow {T_n} = 1 + \dfrac{{3{n^2} - 3n - 2n + 2}}{2} \\\ \Rightarrow {T_n} = 1 + \dfrac{{3{n^2} - 5n + 2}}{2} \\\$
On solving further we take $2$ as LCM, we get
$\Rightarrow {T_n} = \dfrac{{2 + 3{n^2} - 5n + 2}}{2} \\\ \Rightarrow {T_n} = \dfrac{{3{n^2} - 5n + 4}}{2} \\\$
The expression can be written as
$\Rightarrow {T_n} = \dfrac{1}{2}\left[ {3{n^2} - 5n + 4} \right]$
Hence this is the ${n^{th}}$ term of given series.

Note: In this type of problem which is based on the series. If terms of any series are not an A.P. but the difference of two consecutive terms of this series is in A.P. then the sum of such series can be found out by Method of Difference.