CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Sum to n terms of the series log m+log m2n+log m3n2+log m4n3 is

A
log(mn+1nn1)n2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
log(nn1mn+1)n2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
log(mnnn)n2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
log(m1nn1m)n2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A log(mn+1nn1)n2
log (m × m2 × m3 × .........mn)log(1 × n × n2 × ......n(n1)) = log[m(n+1)n(n1)](n/2)

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Series
QUANTITATIVE APTITUDE
Watch in App
Join BYJU'S Learning Program
CrossIcon