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If $a=\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ and $b=\lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$, then
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The correct answer is:
$2 \mathrm{a}=3 \mathrm{~b}$
$$
\begin{aligned}
& a=\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2} \\
& =\lim _{n \rightarrow \infty} \frac{n(n+1)}{2(n)(n)}=\lim _{n \rightarrow \infty}\left(\frac{1}{2}\right)\left(\frac{n+1}{n}\right)=\frac{1}{2} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=\frac{1}{2} \\
& b=\lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots .+n^2}{n^3} \\
& =\lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^3}=\lim _{n \rightarrow \infty} \frac{2 n^2+3 n+1}{6 n^2} \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{6}\right)\left(2+\frac{3}{n}+\frac{1}{n^2}\right)=\frac{2}{6}=\frac{1}{3}
\end{aligned}
$$
Thus $\mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=\frac{1}{3} \Rightarrow 2 \mathrm{a}=3 \mathrm{~b}$
\begin{aligned}
& a=\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2} \\
& =\lim _{n \rightarrow \infty} \frac{n(n+1)}{2(n)(n)}=\lim _{n \rightarrow \infty}\left(\frac{1}{2}\right)\left(\frac{n+1}{n}\right)=\frac{1}{2} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=\frac{1}{2} \\
& b=\lim _{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots .+n^2}{n^3} \\
& =\lim _{n \rightarrow \infty} \frac{n(n+1)(2 n+1)}{6 n^3}=\lim _{n \rightarrow \infty} \frac{2 n^2+3 n+1}{6 n^2} \\
& =\lim _{n \rightarrow \infty}\left(\frac{1}{6}\right)\left(2+\frac{3}{n}+\frac{1}{n^2}\right)=\frac{2}{6}=\frac{1}{3}
\end{aligned}
$$
Thus $\mathrm{a}=\frac{1}{2}$ and $\mathrm{b}=\frac{1}{3} \Rightarrow 2 \mathrm{a}=3 \mathrm{~b}$
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