JEE Series

This page will teach you how to master JEE Series. We highlight the main concepts, provide a list of examples with solutions, and include problems for you to try. Once you are confident, you can take the quiz to establish your mastery.

As per JEE syllabus, the main concepts under series are $\sum n , \sum n^2 \text{ and } \sum n^3$ series, arithmetic-geometric series, telescopic sum, telescopic product and successive difference.

$\sum n , \sum n^2 \text{ and } \sum n^3$ series

• Sum of first $n$ natural numbers, abbreviated as $\sum n$, is given by $\sum n=1+2+3+...+n=\frac{n(n+1)}{2}$

• Sum of squares of first $n$ natural numbers, abbreviated as $\sum n^2$, is given by $\sum n^2=1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

• Sum of cubes of first $n$ natural numbers, abbreviated as $\sum n^3$, is given by $\sum n^3=1^3+2^3+3^3+...+n^3=\left( \frac{ n(n+1)}{2} \right)^2$

Arithmetic-Geometric Progression (A.G.P.)

• Sum of first $n$ terms of A.G.P. is given by $a+(a+d)r+(a+2d)r^2+(a+3d)r^3+...+\left( a+(n-1)d \right) r^{n-1}$ $=\frac{a}{1-r} +\frac{dr}{(1-r)^2} -\frac{dr^n}{(1-r)^2}- \frac{\left( a+(n-1)d \right) r^n}{1-r}$

• Sum of infinite terms of A.G.P. is given by $S_{\infty} =\frac{a}{1-r} +\frac{dr}{(1-r)^2}$

Telescopic sum and product

• Telescopic sum also known as method of difference is used when the general term of the series can be expressed as the difference of two terms by clever manipulation, as $t_r=f(r)-f(r \pm 1)$

• Telescopic product

Successive difference

• Finding the $n$th term using "Successive difference" of second order A.P.

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Once you are confident of JEE Series, move on to JEE Matrices and Determinants.