If n=2006! then 1log2n+1log3n+....+1log2006n=
2006
2005
2005!
1
0
Explanation for correct option
Given n=2006!
∴1log2n+1log3n+....+1log2006n=logn2+logn3+..+logn2006∵1logab=logba=logn2·3·4·...·2006∵logab+logac=logab·c=log2006!2006!∵n=2006!=1∵logaa=1
Hence, option(D) i.e. 1 is correct
Determine whether the following numbers are in proportion or not:
13,14,16,17
loge(n+1)−loge(n−1)=4a[(1n)+(13n3)+(15n5)+...∞] Find 8a.