JEE Main & Advanced Mathematics Sequence & Series Method of Difference

If the differences of the successive terms of a series are in A.P. or G.P., we can find \[{{n}^{th}}\] term of the series by the following steps :

Step I: Denote the \[{{n}^{th}}\] term by \[{{T}_{n}}\] and the sum of the series upto \[n\] terms by \[{{S}_{n}}\].

Step II: Rewrite the given series with each term shifted by one place to the right.

Step III: By subtracting the later series from the former, find \[{{T}_{n}}\].

Step IV: From \[{{T}_{n}}\], \[{{S}_{n}}\] can be found by appropriate summation.

Example : Consider the series 1+ 3 + 6 + 10 + 15 +…..to \[n\] terms. Here differences between the successive terms are \[63,\text{ }106,\text{ }1510,\text{ }\ldots \ldots .\] i.e.,  2, 3, 4, 5,…… which are in A.P. This difference could be in G.P. also. Now let us find its sum

\[S=1+3+6+10+15+.....+{{T}_{n-1}}+{{T}_{n}}\]

\[S=\,\,\,\,\,\,\,\,\,1+3+6+10+..........+{{T}_{n-1}}+{{T}_{n}}\]

Subtracting, we get

\[0=1+2+3+4+5+.........+({{T}_{n}}-{{T}_{n-1}})-{{T}_{n}}\]

\[\Rightarrow \]      \[{{T}_{n}}=1+2+3+4+.........\]to \[n\] terms.

\[\Rightarrow \]      \[{{T}_{n}}=\frac{1}{2}n(n+1)\]     \[\therefore \] \[{{S}_{n}}=\Sigma {{T}_{n}}=\frac{1}{2}[\Sigma {{n}^{2}}+\Sigma n]\]

= \[\frac{1}{2}\left[ \frac{n(n+1)\,(2n+1)}{6}+\frac{n\,(n+1)}{2} \right]\] = \[\frac{n\,(n+1)\,(n+2)}{6}\].