The value of ∑r=1nlogarbr-1 is
n2loganbn
n2logan+1bn
n2logan+1bn-1
n2logan+1bn+1
The explanation for the correct option
The given expression: ∑r=1nlogarbr-1.
∑r=1nlogarbr-1=loga1b0+loga2b1+....+loganbn-1⇒∑r=1nlogarbr-1=loga1b0×a2b1×......×anbn-1∵logAB=logA+logB⇒∑r=1nlogarbr-1=loga1+2+....+nb1+2+....+n-1
Now, the sum of the first n natural can be given by, Sn=nn+12.
we know that 1+2+.....+n=nn+12.
And, 1+2+.....+n-1=n-1n-1+12=nn-12.
Thus, ∑r=1nlogarbr-1=loga1+2+....+nb1+2+....+n-1
⇒∑r=1nlogarbr-1=logann+12bnn-12⇒∑r=1nlogarbr-1=logan+1bn-1n2⇒∑r=1nlogarbr-1=n2logan+1bn-1∵logAa=alogA
Therefore, the value of ∑r=1nlogarbr-1 is n2logan+1bn-1.
Hence, the correct option is (C).