The greatest value of n for which 1+12+122+.....+12n<2 is (n∈N).
100
10
1000
None of these
The explanation for the correct option.
Find the greatest value.
Given 1+12+122+.....+12n<2
Sum of GP =a(1-rn)1-r
=11-12n+11-12=21-12n+1=2-12n<2
Hence, option D is the correct answer.
loge(n+1)−loge(n−1)=4a[(1n)+(13n3)+(15n5)+...∞] Find 8a.
Determine whether the following numbers are in proportion or not:
13,14,16,17