Tag: maths

$1+3+7+15+31+.....$ to n terms 

If $1+a+a^{2}+a^{3}+.........+a^{n}=(1+a)(1+a^2)(1+a^4)$ then $n$ is given by 

The sum of first 4 term of GP with $a=2,r=3$ is 

Three numbers whose sum is $45$ are in A.P. If $5$ is subtracted from the first number and $25$ is added to third number, the numbers are in G.P. Then numbers can be

If $S$ is the sum to infinity of a $G.P.$ whose first terms is $a$ then the sum of the first $n$ terms is 

For first $n$ natural numbers we have the following results with usual notations $ \displaystyle \sum _{r=1}^{n}r =\frac{n(n+1)}{2}, \sum _{r=1}^{n}r^{2} =\frac{n(n+1)(2n+1)}{6},\sum _{r=1}^{n}r^{3}=\left ( \sum _{r=1}^{n}r \right )^{2}$ If $\displaystyle a _{1}a _{2}....a _{n} \in A.P $ then sum to $n$ terms of the sequence $\displaystyle \frac{1}{a _{1}a _{2}},\frac{1}{a _{2}a _{3}},...\frac{1}{a _{n-1}a _{n}}$ is equal to $\displaystyle \frac{n-1}{a _{1}a _{n}}$
 and the sum to $ n$ terms of a $G.P$ with first term '$a$' & common ratio '$r$' is given by  $\displaystyle S _{n}= \frac{lr-a}{r-1}$ for $ r \neq 1 $ for $ r =1 $ sum to $n$ terms of same $G.P.$ is $n$ $a$, where the sum to infinite terms of$G.P.$ is the limiting value of
 $\displaystyle \frac{lr-a}{r-1} $ when $\displaystyle n \rightarrow \infty ,\left |  r \right | < l $ where $l$ is the last term of $G.P.$  On the basis of above data answer the following questionsThe sum of the series $\displaystyle 2+6+18+...+486 $ equals?

$\displaystyle x(x+y)+x^{2}(x^{2}+y^{2})+x^{3}(x^{3}+y^{3})+$.......to n terms.

Find the value of the sum $\displaystyle \sum _{r=1}^{n}\,$ $\displaystyle \sum _{s=1}^{n}\, \delta _{rs}\, 2^r\, 3^s$ where $ \delta _{rs}$ is zero if $r \neq s$ & $\delta _{rs}$ is one if $r=s$

The A.M. of the series $1, 2, 4, 8, 16, ......, 2$$^n$ is

$\displaystyle 1+\frac{1}{4\times 3}+\frac{1}{4\times 3^{2}}+\frac{1}{4\times 3^{3}}$  is equal to