The Sum Of The Series 2 C0+C1/2· 22+C2/3· 23+C3/4· 24+⋯+Cn/n+1

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The sum of the series 2C0+C12⋅22+C23⋅23+C34⋅24+⋯+Cnn+1⋅2n+1 is equal to , where(Cr=nCr)

A3n+1−1n−1No worries! We‘ve got your back. Try BYJU‘S free classes today!B3n+1−1n+1Right on! Give the BNAT exam to get a 100% scholarship for BYJUS coursesC3n+1+1n+1No worries! We‘ve got your back. Try BYJU‘S free classes today!D3n−1−1n+1No worries! We‘ve got your back. Try BYJU‘S free classes today!Open in AppSolution

The correct option is B 3n+1−1n+1We have, 2C0+C12⋅22+C23⋅23+C34⋅24+⋯+Cnn+1⋅2n+1 =n∑r=01r+1nCr⋅2r+1 =1n+1n∑r=0n+1r+1nCr⋅2r+1 =1n+1n∑r=0n+1Cr+1⋅2r+1 =1n+1n+1∑s=1n+1Cs⋅2s[Puttingr+1=s] =1n+1[(n+1∑s=0n+1Cs2s)−n+1C0⋅20] =1n+1[(1+2)n+1−1]=3n+1−1n+1

flagSuggest Correctionsthumbs-up65similar_iconSimilar questionsQ. C0+32.C1+C2+274.C3++3nn+1.Cn=4n+1+13(n+1)Q. If (1+x)n=C0+C1x+C2x2+...+Cnxn, then 2C0+22.C12+23.C23+...+2n+1.Cnn+1=Q. The sum to (n+1) terms of the following series C02−C13+C24−C35+⋯ isQ. Let S=21nC0+222nC1+233nC2+⋯+2n+1n+1nCn. Then, S equalsQ. Let S=21nC0+222nC1+233nC2+.....+2n+1n+1nCn. Then S equalsView MoreJoin BYJU'S Learning ProgramGrade/Exam1st Grade2nd Grade3rd Grade4th Grade5th Grade6th grade7th grade8th Grade9th Grade10th Grade11th Grade12th GradeSubmitexplore_more_iconExplore moreSum of Product of Binomial CoefficientsStandard XII MathematicsJoin BYJU'S Learning ProgramGrade/Exam1st Grade2nd Grade3rd Grade4th Grade5th Grade6th grade7th grade8th Grade9th Grade10th Grade11th Grade12th GradeSubmitbook_imageAI Tutorbook_imageTextbooksbook_imageQuestion Papersbook_imageInstall appCrossIcon

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