Saturday, 1 October 2016

sequences and series - Prove that frac1a1a2+frac1a2a3+frac1a3a4+...+frac1an1an=fracn1a1an




There is this question in one of my math textbooks which I can't seem to figure out how to solve, it'd be awesome if you could help me :




If a1,a2,a3,...,an IS an arithmetic progression and an is NOT equal to 0 then prove the following statement :



1a1a2+1a2a3+1a3a4+...+1an1an=n1a1an




Thanks in advance...



Answer



Here's sketch, not complete proof.



Since A is an arithmetic progression, Let ai+1ai=d



1aiai+1=dd1aiai+1=1dai+1aiaiai+1since ai+1ai=d=1d(1ai1ai+1)



Therefore, 1a1a2+1a2a3++1an1an=n1i=11aiai+1=1dn1i=11ai1ai+1=1d(1a11a2+1a21an1+1an11an)=1d(1a11an)



Now, take LCM and use the fact that an=a1+(n1)d


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