Question Number 20523 by Tinkutara last updated on 27/Aug/17
$$\mathrm{Let}\:\mathrm{the}\:\mathrm{sum}\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\frac{\mathrm{1}}{{n}\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}\:\mathrm{written} \\ $$$$\mathrm{in}\:\mathrm{its}\:\mathrm{lowest}\:\mathrm{terms}\:\mathrm{be}\:\frac{{p}}{{q}}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:{q}\:−\:{p}. \\ $$
Answered by ajfour last updated on 27/Aug/17
![S=(1/2)Σ_(n=1) ^9 (((n+2)−n)/(n(n+1)(n+2))) =(1/2)Σ_(n=1) ^9 [(1/(n(n+1)))−(1/((n+1)(n+2)))] =(1/2)[((1/(1.2))−(1/(2.3)))+((1/(2.3))−(1/(3.4)))+.... ....+((1/(9.10))−(1/(10.11)))] =(1/2)[(1/2)−(1/(110))] = ((27)/(110)) ⇒ (p/q) = ((27)/(110)) q−p= 83 .](Q20525.png)
$${S}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\frac{\left({n}+\mathrm{2}\right)−{n}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\left[\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}}\right)+....\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....+\left(\frac{\mathrm{1}}{\mathrm{9}.\mathrm{10}}−\frac{\mathrm{1}}{\mathrm{10}.\mathrm{11}}\right)\right] \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{110}}\right]\:=\:\frac{\mathrm{27}}{\mathrm{110}} \\ $$$$\Rightarrow\:\:\:\frac{{p}}{{q}}\:=\:\frac{\mathrm{27}}{\mathrm{110}} \\ $$$$\:\boldsymbol{{q}}−\boldsymbol{{p}}=\:\mathrm{83}\:. \\ $$
Commented by Tinkutara last updated on 28/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$