Method of Differences for Finding Sum

Let the sum ${\displaystyle \sum _{n=1}^9\frac{1}{n\left(n+1\right)\left(n+2\right)}}$ , written in the rational form be $\frac{p}{q}$ (where $p$ and $q$ are co-prime), then the value of $\left[\frac{q-p}{10}\right]$ is, (where [.] is the greatest integer function)

For any positive integer $n$ let $f\left(n\right)=\frac{4n+\sqrt{4n^2–1}}{\sqrt{2n+1}+\sqrt{2n–1}}$ and $\sum _{K=1}^{n}f\left(K\right)=\alpha \left({\left(2n+1\right)}^{\beta }-1\right)$, then

The sum of the series $\frac{2}{1·2}+\mathrm{ }\frac{5}{2·3}·2\mathrm{ }+\mathrm{ }\frac{10}{3·4}·2^2\mathrm{ }+\mathrm{ }\frac{17}{4·5}·2^3\mathrm{ }+$ ...... to $\mathrm{n}$ terms is

If $\underset{k=1}{\overset{\infty }{\Sigma }}\frac{1}{\left(k+2\right)\sqrt{k}+k\sqrt{k+2}}=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{c}}$, where $a, b, c\in N$ and $a, b, c\in [1, 15]$, then $a+ b+c$ is equal to

Value of series $\sum _{n=1}^{\infty }\left(\frac{n}{n^4 + 4}\right)$ is equal to

If $\mathrm{S}=\frac{5}{1^2·2^2·3}+\frac{8}{2^2·3^2·4}+\frac{11}{3^2·4^2·5}+\frac{14}{4^2·5^2·6}+...$, then $4\mathrm{S}$ is:

Let $S=\sum _{n=1}^{\infty }\frac{n}{n^4+4}$, then the value of $144S$ is

If the sum of $n$ terms of the series $\frac{5}{1.2.3}+\frac{6}{2.3.4}+\frac{7}{3.4.5}+\dots \dots \dots$ is $\frac{a}{2}-\frac{n+b}{(n+1)(n+2)},$ then :

The sum of the series $1+\frac{1}{3^2}+\frac{1}{1}\cdot \frac{4}{2}\cdot \frac{1}{3^4}+\frac{1}{1}\cdot \frac{4}{2}\cdot \frac{7}{2}\cdot \frac{1}{3^6}+\dots ..$ is

The value of $\sum _{n=1}^{\infty }\frac{n}{n^4+4}$ is equal to

The sum of the infinite series

$1+\left(1+\frac{1}{2}\right)\left(\frac{1}{3}\right)+ \left(1+\frac{1}{2}+\frac{1}{2^2}\right)\left(\frac{1}{3^2}\right)+\dots \infty$ is

If $t_n=\frac{1}{4}\left(n+2\right)\left(n+3\right)$ for $n=1,\mathrm{2,3},\dots .$ Then $\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\dots +\frac{1}{t_{2003}}=$

The series $1\mathrm{ }.\mathrm{ }1!\mathrm{ }+\mathrm{ }2\mathrm{ }.\mathrm{ }2!\mathrm{ }+\mathrm{ }3\mathrm{ }.\mathrm{ }3!\mathrm{ }+\mathrm{ }\dots ..\mathrm{ }+\mathrm{ }n\mathrm{ }.\mathrm{ }n!\mathrm{ }=$

$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+2^2}{3!}+\dots \infty =$

If ${\mathrm{t}}_{\mathrm{n}}=\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)$, then $\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{t}}_{\mathrm{n}}}$ is equal to

The sum of the series $\frac{2}{1.2}+\frac{5}{2.3}.2+\frac{10}{3.4}.2^2+\frac{17}{4.5}.2^3+\dots$ to $n$ terms is 

The sum to $n$ terms of the series

$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\dots +\frac{\mathrm{n}}{1+{\mathrm{n}}^2+{\mathrm{n}}^4}$ is

The sum of $\frac{3}{1.2}.\frac{1}{2}+\frac{4}{2.3}{\left(\frac{1}{2}\right)}^2+\frac{5}{3.4}.{\left(\frac{1}{2}\right)}^3+\dots$ to $n$ terms, is equal to

Sum to n terms $\left[\frac{1}{1.3}+\frac{2}{1.3.5}+\frac{3}{1.3.5.7}+\frac{4}{1.3.5.7.9}+\dots \mathrm{ }\right]$ is

If ${\mathrm{a}}_1,\mathrm{ }{\mathrm{a}}_2,\mathrm{ }{\mathrm{a}}_3\dots .\mathrm{ }$ are in $A.P$ and ${\mathrm{a}}_1>0$ for each I, then $\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{\mathrm{n}}{{\mathrm{a}}_{\mathrm{i}+1}^{\frac{2}{3}}+{\mathrm{a}}_{\mathrm{i}+1}^{\frac{1}{3}}.{\mathrm{a}}_{\mathrm{i}}^{\frac{1}{3}}+{\mathrm{a}}_{\mathrm{i}}^{\frac{2}{3}}}$ is equal to