A) 5
B) 2
C) 1
D) 0
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: \[1+2+3+...+n=\frac{n(n+1)}{2}\] \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{n}^{2}}=\frac{n(n+1)(2n+1)}{6}\]
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