find the value of log m + log m2 + log m3 + ...... + log mn : a) (n(n+1)/2 b) mn/2 c) (n(n+1)) log m/2 d) n(n+1) log m2
Answers
Answered by
2
The value of (logm+logm
2
+logm
3
+...+logm
n
) is equal to
Share
Study later
ANSWER
logm+logm
2
+logm
3
+...+logm
n
=log(m.m
2
.m
3
...m
n
)
=logm
(1+2+3+...n)
=logm
2
n(n+1)
=
2
n(n+1)
logm
I
Similar questions