Limn→∞1n4[12+(12+22)+.....(12+22+.......+n2)]=

limn→∞∑n(n+1)(2n+1)6n4 =limn→∞∑ 2n3+3n2+n6n4 =limn→∞2n2(n+1)24+3n(n+1)(2n+1)6+n(n+1)2n4=24=12

Limn→∞1n4[12+(12+22)+.....(12+22+.......+n2)]=

limn→∞∑n(n+1)(2n+1)6n4 =limn→∞∑ 2n3+3n2+n6n4 =limn→∞2n2(n+1)24+3n(n+1)(2n+1)6+n(n+1)2n4=24=12

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$\underset{n \to \infty}{L i m} \frac{1}{n^{4}} [1^{2} + (1^{2} + 2^{2}) + ..... (1^{2} + 2^{2} + ....... + n^{2})] =$

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Detailed Solution

$\underset{n \to \infty}{l i m} \frac{\sum \frac{n (n + 1) (2 n + 1)}{6}}{n^{4}} = \lim_{n \to \infty} \frac{\sum_{} 2 n^{3} + 3 n^{2} + n}{6 n^{4}} = \lim_{n \to \infty} \frac{\frac{2 n^{2} {(n + 1)}^{2}}{4} + \frac{3 n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}}{n^{4}} = \frac{2}{4} = \frac{1}{2}$