The Calculator Just Surrendered!

We have to find the sum of the sequence $\frac{1}{7.8.9}+\frac{1}{8.9.10}+........+\frac{1}{20.21.22}$

Now, We clearly see that we have to make use of telescoping series. So how can we use it?, lets see

first, the general term of the sequence is given by $a_n=\frac{1}{(n+6)(n+7)(n+8)}$ So, we see that $\frac{1}{a_n}=(n+6)(n+7)(n+8)$ Similarly $\frac{1}{a_{n-1}}=(n+5)(n+6)(n+7)$ Now, by the above two results, we get $\frac{a_{n-1}}{a_n}=\frac{n+8}{n+5}$ $\Rightarrow (n+5)a_{n-1}=(n+8)a_n$ Let $(n+5)a_{n-1}=(n+8)a_n=b_n$ So, we have $b_n=(n+5)a_{n-1}$ Now, put $n+1$ in the above result, and we obtain ;

$b_{n+1}=(n+6)a_n$

Now, we have $b_n=(n+8)a_n , \; \; \; b_{n+1}=(n+6)a_n$ So, subtract these above terms, and we get ;

$b_n-b_{n+1} = 2 \times a_n$

So, we have got a good looking result now. So, lets put $n=1,2,...,14$ , and add, so we get

$b_1-b_2+b_2-b_3+.........+b_{14}-b_{15}=2(a_1+a_2+....a_{14})=2\times S_{14}.........(1)$

Note that $S_{14}$ is what is asked to us. So, in equation $(1)$ , we have telescoping series on LHS. and we get

$2 \times S_{14}=b_1-b_{15}$

Now, $b_1=(1+8)\times a_1=\frac{1}{7.8},b_{15}=(15+8)\times a_{15}=\frac{1}{21.22}$

So, $2 \times S_{14}=\frac{29}{1848}$ $\Rightarrow S_{14}=\frac{29}{3696}$

This is a telescopic series using partial fractions.
$Its~ general~ term~ is ~\dfrac{1}{(n-1)n(n+1)}$ $its~ partial ~fractions ~are:-\dfrac{1/2}{n - 1} - \dfrac{1}{n}+ \dfrac{1/2}{n + 1} ~\\~$ $So~ the ~SUM~ is ~= \sum_8^{21} \{ \dfrac{1/2}{n - 1} - \dfrac{1}{n}+\dfrac{1/2}{n + 1} \} \\~$

$\\ \sum_8^{21}\dfrac{\frac{1}{2}}{n - 1} =\sum_7^{20}\dfrac{\frac{1}{2}}{n} = \sum_7^{8}\dfrac{1}{2n} +\sum_9^{20}\dfrac{1}{2n} =\dfrac{1}{2*7} +\dfrac{1}{2*8}+\sum_9^{20}\dfrac{1}{2n}....(1) \\~\\$ $\\~\\$ $\sum_8^{21}\dfrac{1}{n} = \sum_8^{8}\dfrac{1}{n}+ \sum_9^{20}\dfrac{1}{n} +\sum_{21}^{21}\dfrac{1}{n}=\dfrac{1}{8} +\sum_9^{20}\dfrac{1}{n}+\dfrac{1}{21}.......(2) \\~\\$

$\\ \sum_8^{21}\dfrac{\frac{1}{2}}{n+ 1} =\sum_9^{22}\dfrac{\frac{1}{2}}{n} = \sum_9^{20}\dfrac{1}{2n} +\sum_{21}^{22}\dfrac{1}{2n}=\sum_9^{20}\dfrac{1}{2n}+\dfrac{1}{2*21} +\dfrac{1}{2*22}....(3) \\~\\$ $\\~\\$

$So~ the~ SUM = (1)~ -~ (2) + (3)~\\$ $~for~ all~ three~\sum_9^{20}\{\dfrac{1}{2n} -\dfrac{1}{n} +\dfrac{1}{2n}\}=0 \\$

$\therefore~\sum=\dfrac{1}{14}+\dfrac{1}{16}-\dfrac{1}{8}- \dfrac{1}{21}+\dfrac{1}{42}+\dfrac{1}{44} ~ =\dfrac{29}{3696}=\dfrac{a}{b}…\\\ a+b= \boxed{3725}$ ,

Each term has this pattern
$\frac{1}{\left(n+6\right)\left(n+7\right)\left(n+8\right)}$

Decompose this to partial fraction.

$\frac{1}{\left(n+6\right)\left(n+7\right)\left(n+8\right)}=\frac{A}{n+6}+\frac{B}{n+7}+\frac{C}{n+8}$

$1=A\left(n+7\right)\left(n+8\right)+B\left(n+6\right)\left(n+8\right)+C\left(n+6\right)(n+7)$

When $n=-6\ ; 1=A\left(1\right)\left(2\right) ; A=\frac{1}{2}$ When $n=-7;; 1=B\left(-1\right)(1); B=-1$ When $n=-8; 1=C\left(-2\right)(-1); C=\frac{1}{2}$

So,
$\frac{1}{\left(n+6\right)\left(n+7\right)\left(n+8\right)}=\frac{1}{2(n+6)}-\frac{1}{n+7}+\frac{1}{2(n+8)}$

or $\frac{1}{\left(n+6\right)\left(n+7\right)\left(n+8\right)}=\frac{1}{2}\left(\frac{1}{n+6}-\frac{2}{n+7}+\frac{1}{n+8}\right)$

Replace now each term in the problem with $\frac{1}{2}\left(\frac{1}{n+6}-\frac{2}{n+7}+\frac{1}{n+8}\right)$

$\frac{1}{7\cdot 8\cdot 9}+\frac{1}{8\cdot 9\cdot 10}+\frac{1}{9\cdot 10\cdot 11}+\dots +\frac{1}{20\cdot 21\cdot 22}$ $=\frac{1}{2}\left(\frac{1}{7}-\frac{2}{8}++\frac{1}{9}+\frac{1}{8}-\frac{2}{9}+\frac{1}{10}+\frac{1}{9}-\frac{2}{10}+\frac{1}{11}+\dots +\frac{1}{18}-\frac{2}{19}+\frac{1}{20}+\frac{1}{19}-\frac{2}{20}+\frac{1}{21}+\frac{1}{20}-\frac{2}{21}+\frac{1}{22}\right)$

By telescoping technique

$=\frac{1}{2}\left(\frac{1}{7}-\frac{1}{8}-\frac{1}{21}+\frac{1}{22}\right)$ $=\frac{1}{2}\left(\frac{1}{7}-\frac{1}{8}-\frac{1}{21}+\frac{1}{22}\right)$ $=\frac{1}{2}\left(\frac{1}{56}-\frac{1}{462}\right)$ $=\frac{1}{2}\left(\frac{33-4}{1848}\right)$ $=\frac{29}{3696}=\frac{\alpha }{\beta }$

$\alpha +\beta =29+3696=3725$

\quad \sum { \frac { 1 }{ x(x+1)(x+2) } =\frac { x(x+3) }{ 4(x+2)(x+3) } } But how is Lorax related to the problem? Satvik Golechha

1/7.8.9 =1/2(1/7.8 -1/8.9) so 1/7.8.9 + ................. +1/20.21.22 =1/2 (1/7.8-1/8.9+1/8.9 -1/9.10...........+1/20.21 -1/21.22 ) =1/2 (1/7.8-1/21.22) =29/3696 so answer is 29+3696 =3725

Here's the shortcut for finding the summation of these.

$\displaystyle \frac{1}{9-7}\left(\frac{1}{7\times8} - \frac{1}{21\times22}\right) = \frac{29}{3696}$

First, we find the difference between the last number and the first number of the same term (any term you want).

Then we take the first term and cancel the last number.

And then we take the last term and cancel the first term.

The general proof is actually the same idea to the cutie rabbit's solution. I've done proving it once, but it took me 3 hours, no time for me to write the whole thing down here. Sorry about that. =="