Differentiation Question 503
Question: What is $ \underset{n\to \infty }{\mathop{\lim }},\frac{1+2+3+…+n}{1^{2}+2^{2}+3^{2}+…n^{2}} $ equal to?
Options:
A) 5
B) 2
C) 1
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ \underset{n\to \infty }{\mathop{\lim }},\frac{1+2+3+…+n}{1^{2}+2^{2}+3^{3}+…+n^{2}} $ $ =\underset{n\to \infty }{\mathop{\lim }},\frac{\frac{n(n+1)}{2}}{\frac{n(n+1)(2n+1)}{6}} $
$ \therefore \underset{n\to \infty }{\mathop{\lim }},\frac{3}{2n+1}=0 $ Note: $ 1+2+3+…+n=\frac{n(n+1)}{2} $ $ 1^{2}+2^{2}+3^{2}+…+n^{2}=\frac{n(n+1)(2n+1)}{6} $
BETA
