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Question Number 2481 by Syaka last updated on 21/Nov/15

((2014)/(2013)) + ((2014)/(2013)) ∗((2012)/(2011)) + ((2014)/(2013)) ∗ ((2012)/(2011)) ∗ ((2010)/(2009)) + ∙∙∙∙ + ((2014)/(2013)) ∗ ((2012)/(2011)) ∗ ((2010)/(2009)) ∗ ((2008)/(2007 ))∗∙∙∙∗(4/3)∗(2/1) − 1 = ?

$$\frac{\mathrm{2014}}{\mathrm{2013}}\:+\:\frac{\mathrm{2014}}{\mathrm{2013}}\:\ast\frac{\mathrm{2012}}{\mathrm{2011}}\:+\:\frac{\mathrm{2014}}{\mathrm{2013}}\:\ast\:\frac{\mathrm{2012}}{\mathrm{2011}}\:\ast\:\frac{\mathrm{2010}}{\mathrm{2009}}\:+\:\centerdot\centerdot\centerdot\centerdot\:+\:\frac{\mathrm{2014}}{\mathrm{2013}}\:\ast\:\frac{\mathrm{2012}}{\mathrm{2011}}\:\ast\:\frac{\mathrm{2010}}{\mathrm{2009}}\:\ast\:\frac{\mathrm{2008}}{\mathrm{2007}\:}\ast\centerdot\centerdot\centerdot\ast\frac{\mathrm{4}}{\mathrm{3}}\ast\frac{\mathrm{2}}{\mathrm{1}}\:−\:\mathrm{1}\:=\:\:\:\:? \\ $$

Answered by Filup last updated on 21/Nov/15

if by ′′∗′′ you mean ′′×′′ in form (k=2014): S=((k/(k−1))+((k(k−2))/((k−1)(k−3)))+((k(k−2)(k−4))/(k(k−1)(k−3)(k−5)))+...)−1 According to WolframAlpha: Sum=(Σ_(n=1) ^m (k/((1+k−2n))))−1 where m=((2014)/(2013))×...(4/3)×(2/1) m=Π_(l=1) ^k ((l+1)/l)=k+1=2015 (W.Alpha) ∴S=Σ_(n=1) ^(2015) (((2014)/(2015−2n)))−1 continue not sure how to evaluate this, if at all correct

$$\mathrm{if}\:\mathrm{by}\:''\ast''\:\mathrm{you}\:\mathrm{mean}\:''×'' \\ $$$${in}\:{form}\:\left({k}=\mathrm{2014}\right): \\ $$$$ \\ $$$${S}=\left(\frac{{k}}{{k}−\mathrm{1}}+\frac{{k}\left({k}−\mathrm{2}\right)}{\left({k}−\mathrm{1}\right)\left({k}−\mathrm{3}\right)}+\frac{{k}\left({k}−\mathrm{2}\right)\left({k}−\mathrm{4}\right)}{{k}\left({k}−\mathrm{1}\right)\left({k}−\mathrm{3}\right)\left({k}−\mathrm{5}\right)}+...\right)−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{According}\:\mathrm{to}\:\mathrm{WolframAlpha}: \\ $$$${Sum}=\left(\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{{k}}{\left(\mathrm{1}+{k}−\mathrm{2}{n}\right)}\right)−\mathrm{1} \\ $$$$\mathrm{where}\:{m}=\frac{\mathrm{2014}}{\mathrm{2013}}×...\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{2}}{\mathrm{1}} \\ $$$${m}=\underset{{l}=\mathrm{1}} {\overset{{k}} {\prod}}\frac{{l}+\mathrm{1}}{{l}}={k}+\mathrm{1}=\mathrm{2015}\:\:\:\:\:\:\:\left({W}.{Alpha}\right) \\ $$$$ \\ $$$$\therefore{S}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2015}} {\sum}}\left(\frac{\mathrm{2014}}{\mathrm{2015}−\mathrm{2}{n}}\right)−\mathrm{1} \\ $$$$ \\ $$$${continue} \\ $$$$\mathrm{not}\:\mathrm{sure}\:\mathrm{how}\:\mathrm{to}\:\mathrm{evaluate}\:\mathrm{this},\:\mathrm{if}\:\mathrm{at}\:\mathrm{all} \\ $$$$\mathrm{correct} \\ $$

Answered by prakash jain last updated on 24/Nov/15

Summing in reverse order last 2 terms excluding −1 ((2014)/(2013))×((2012)/(2011))×...×(4/3)×(2/1)+((2014)/(2013))×((2012)/(2011))×...×(4/3) =((2014)/(2013))×((2012)/(2011))×...×(4/3)((2/1)+1)=((2014)/(2013))×((2012)/(2011))×...(6/5)×4 ((2014)/(2013))×((2012)/(2011))×...(6/5)×4+((2014)/(2013))×((2012)/(2011))×...×(6/5)=((2014)/(2013))×((2012)/(2011))×...6 Continuing in the same way ((2014)/(2013))+((2014)/(2013))×((2012)/(2011))+...+((2014)/(2013))×((2012)/(2011))×((2010)/(2009))×..×(2/1) = 2014 Final sum=2014−1=2013

$$\mathrm{Summing}\:\mathrm{in}\:\mathrm{reverse}\:\mathrm{order}\:\mathrm{last}\:\mathrm{2}\:\mathrm{terms}\:\mathrm{excluding}\:−\mathrm{1} \\ $$$$\frac{\mathrm{2014}}{\mathrm{2013}}×\frac{\mathrm{2012}}{\mathrm{2011}}×...×\frac{\mathrm{4}}{\mathrm{3}}×\frac{\mathrm{2}}{\mathrm{1}}+\frac{\mathrm{2014}}{\mathrm{2013}}×\frac{\mathrm{2012}}{\mathrm{2011}}×...×\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{2014}}{\mathrm{2013}}×\frac{\mathrm{2012}}{\mathrm{2011}}×...×\frac{\mathrm{4}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\mathrm{1}}+\mathrm{1}\right)=\frac{\mathrm{2014}}{\mathrm{2013}}×\frac{\mathrm{2012}}{\mathrm{2011}}×...\frac{\mathrm{6}}{\mathrm{5}}×\mathrm{4} \\ $$$$\frac{\mathrm{2014}}{\mathrm{2013}}×\frac{\mathrm{2012}}{\mathrm{2011}}×...\frac{\mathrm{6}}{\mathrm{5}}×\mathrm{4}+\frac{\mathrm{2014}}{\mathrm{2013}}×\frac{\mathrm{2012}}{\mathrm{2011}}×...×\frac{\mathrm{6}}{\mathrm{5}}=\frac{\mathrm{2014}}{\mathrm{2013}}×\frac{\mathrm{2012}}{\mathrm{2011}}×...\mathrm{6} \\ $$$$\mathrm{Continuing}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{way} \\ $$$$\frac{\mathrm{2014}}{\mathrm{2013}}+\frac{\mathrm{2014}}{\mathrm{2013}}×\frac{\mathrm{2012}}{\mathrm{2011}}+...+\frac{\mathrm{2014}}{\mathrm{2013}}×\frac{\mathrm{2012}}{\mathrm{2011}}×\frac{\mathrm{2010}}{\mathrm{2009}}×..×\frac{\mathrm{2}}{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2014} \\ $$$$\mathrm{Final}\:\mathrm{sum}=\mathrm{2014}−\mathrm{1}=\mathrm{2013} \\ $$

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