CAT 2003 - RetakeQA Question 46

LogarithmsEasy

What is the sum of n terms in the series

logm+logm2n+logm3n2+logm4n3++logmnnn1?

Answer & solution

Correct answer: log m ( n + 1 ) n ( n - 1 ) n 2

  • A

    log[nn-1m(n+1)]n2

  • B

    log[mmnn]n2

  • C

    log[m(1-n)n(1-m)]n2

  • log[m(n+1)n(n-1)]n2

Solution

logm+log(m2n)+log(m3n2)+log(m4n3)++log(mnnn1)=log[m×m2×m3××mn1×n×n2××nn1]=log[m1+2+3+4++nn0+1+2+3++(n1)]=log[mn2(n+1)nn2(n1)]=log[m(n+1)n(n1)]n2

Hence, option (d).

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CAT 2003 - Retake QA Q46 — Logarithms solution | TheCATExam