Q46 Algebra - Logarithms MULTIPLE CHOICE What is the sum of n terms in the series logm+logm2n+logm3n2+logm4n3+⋯+logmnnn−1? A lognn-1m(n+1)n2 B logmmnnn2 C logm(1-n)n(1-m)n2 D logm(n+1)n(n-1)n2 Show solution Discuss Report an error Save Answer: D logm+logm2n+logm3n2+logm4n3+…+logmnnn−1=logm×m2×m3×…×mn1×n×n2×…×nn−1=logm1+2+3+4+…+nn0+1+2+3+…+(n−1)=logmn2(n+1)nn2(n−1)=logm(n+1)n(n−1)n2 Hence, option (d).