CAT 2003 - Retake — QA Question 46

LogarithmsEasy

What is the sum of n terms in the series

log⁡m+log⁡m2n+log⁡m3n2+log⁡m4n3+⋯+log⁡mnnn−1?

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

log⁡m+log⁡(m2n)+log⁡(m3n2)+log⁡(m4n3)+…+log⁡(mnnn−1)=log⁡[m×m2×m3×…×mn1×n×n2×…×nn−1]=log⁡[m1+2+3+4+…+nn0+1+2+3+…+(n−1)]=log⁡[mn2(n+1)nn2(n−1)]=log⁡[m(n+1)n(n−1)]n2

Hence, option (d).

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CAT 2003 - Retake QA Q46 — Logarithms solution | TheCATExam