Browse · MathNet Print → VII OBM Brazil algebra Problem a, b, c, d are integers with ad=bc. Show that (ax+b)(cx+d)1 can be written in the form ax+br+cx+ds. Find the sum 1⋅41+4⋅71+7⋅101+⋯+2998⋅30011 Solution — click to reveal ax+ba−cx+dc=(ax+b)(cx+d)ad−bc, so we can take r=ad−bca,s=−ad−bcc.k=0∑999(3k+1)(3k+4)1=31k=0∑999(3k+11−3k+41)=31(1−30011)=30011000 Final answer 1000/3001 Techniques Telescoping seriesFractions ← Previous problem Next problem →