![\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|\left(\frac{5n+15}{2n-1}\right)^n\right|}=\lim_{n\to\infty}\frac{5n+15}{2n-1}=\dfrac52](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Csqrt%5Bn%5D%7B%5Cleft%7C%5Cleft%28%5Cfrac%7B5n%2B15%7D%7B2n-1%7D%5Cright%29%5En%5Cright%7C%7D%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B5n%2B15%7D%7B2n-1%7D%3D%5Cdfrac52)
Since this limit exceeds 1, the series diverges.
7.35 in the answer which is what I had got in the calculator ;]
Answer:
17
Step-by-step explanation:
3 caterpillars double every minute (3×2=6). after 4 minutes (6×4=24). He captured 7 caterpillars within the four minutes (24-7=17).
Answer:
23
Step-by-step explanation:
since the number is relatively prime to the product of the first 20 positive numbers
It number must not have factor of (1-20)
Therefore the smallest possible number is the next prime after 20
Answer is 23
Last choice 1 9/50
118/100 =
(2 × 59)/(22 × 52) =
((2 × 59) ÷ 2) / ((22 × 52) ÷ 2) =
59/(2 × 52) =
59/50
Then
59 ÷ 50 = 1 and remainder = 9 =>
59 = 1 × 50 + 9 =>
59/50 =
(1 × 50 + 9) / 50 =
1 + 9/50 =
1 9/50
