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Problem-15-Find-the-sum-of-S-3-1-2-3-4-2-3-4-5-3-4-5-2016-2014-2015-2016-

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Question Number 8554 by Sopheak last updated on 16/Oct/16
Problem .15 Find the sum of S= (3/(1!+2!+3!))+(4/(2!+3!+4!))+(5/(3!+4!+5!))+...+((2016)/(2014!+2015!+2016!))
$${Problem}\:.\mathrm{15} \\ $$$${Find}\:{the}\:{sum}\:{of} \\ $$$${S}=\:\frac{\mathrm{3}}{\mathrm{1}!+\mathrm{2}!+\mathrm{3}!}+\frac{\mathrm{4}}{\mathrm{2}!+\mathrm{3}!+\mathrm{4}!}+\frac{\mathrm{5}}{\mathrm{3}!+\mathrm{4}!+\mathrm{5}!}+…+\frac{\mathrm{2016}}{\mathrm{2014}!+\mathrm{2015}!+\mathrm{2016}!} \\ $$$$\: \\ $$
Commented by Yozzias last updated on 16/Oct/16
(3/(1!(1+2+2×3))),(4/(2!(1+3+3×4))),(5/(3!(1+4+4×5))),... ((n+2)/(n!(1+n+1+(n+1)(n+2)))) =((n+2)/(n!(2+n+n^2 +3n+2))) =((n+2)/(n!(4+4n+n^2 ))) =((n+2)/(n!(n+2)^2 )) =(1/(n!(n+2))) S=Σ_(r=1) ^n (1/(r!(r+2)))=Σ_(r=1) ^n ((r+1)/((r+2)!))=Σ_(r=1) ^(n+1) (r/((r+1)!))−(1/((1+1)!))=1−(1/((n+2)!))−(1/2) S=(1/2)−(1/((n+2)!)) n=2014⇒S=(1/2)−(1/(2016!)) Σ_(r=1) ^(2014) ((r+2)/(r!+(r+1)!+(r+2)!))=(1/2)−(1/(2016!))
$$\frac{\mathrm{3}}{\mathrm{1}!\left(\mathrm{1}+\mathrm{2}+\mathrm{2}×\mathrm{3}\right)},\frac{\mathrm{4}}{\mathrm{2}!\left(\mathrm{1}+\mathrm{3}+\mathrm{3}×\mathrm{4}\right)},\frac{\mathrm{5}}{\mathrm{3}!\left(\mathrm{1}+\mathrm{4}+\mathrm{4}×\mathrm{5}\right)},… \\ $$$$\frac{\mathrm{n}+\mathrm{2}}{\mathrm{n}!\left(\mathrm{1}+\mathrm{n}+\mathrm{1}+\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)\right)} \\ $$$$=\frac{\mathrm{n}+\mathrm{2}}{\mathrm{n}!\left(\mathrm{2}+\mathrm{n}+\mathrm{n}^{\mathrm{2}} +\mathrm{3n}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{n}+\mathrm{2}}{\mathrm{n}!\left(\mathrm{4}+\mathrm{4n}+\mathrm{n}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{n}+\mathrm{2}}{\mathrm{n}!\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{n}!\left(\mathrm{n}+\mathrm{2}\right)} \\ $$$$\mathrm{S}=\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{r}!\left(\mathrm{r}+\mathrm{2}\right)}=\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{r}+\mathrm{1}}{\left(\mathrm{r}+\mathrm{2}\right)!}=\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}+\mathrm{1}} {\sum}}\frac{\mathrm{r}}{\left(\mathrm{r}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{1}\right)!}=\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)!}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{S}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)!} \\ $$$$\mathrm{n}=\mathrm{2014}\Rightarrow\mathrm{S}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2016}!} \\ $$$$\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{2014}} {\sum}}\frac{\mathrm{r}+\mathrm{2}}{\mathrm{r}!+\left(\mathrm{r}+\mathrm{1}\right)!+\left(\mathrm{r}+\mathrm{2}\right)!}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2016}!} \\ $$
Answered by Yozzias last updated on 16/Oct/16
Check answer in comments.
$$\mathrm{Check}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{comments}. \\ $$
Commented by Sopheak last updated on 16/Oct/16
Right answer
$${Right}\:{answer} \\ $$

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