1.If a=limn→∞1+2+3+…+nn2a=\displaystyle\lim_{n\to\infty}\frac{1+2+3+\ldots+n}{n^2}a=n→∞limn21+2+3+…+n and b=limn→∞12+22+…+n2n3b=\displaystyle\lim_{n\to\infty}\frac{1^2+2^2+\ldots+n^2}{n^3}b=n→∞limn312+22+…+n2, thena.a=ba=ba=bb.2a=3b2a=3b2a=3bc.a=2ba=2ba=2bd.3a=2b3a=2b3a=2bLogin to continueOnly logged in users canattempt or see the solution.
