# Current Affairs JEE Main & Advanced

## Method of Difference

Category : JEE Main & Advanced

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.

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

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

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