limit x tends to infinity 1.n+2(n-1)+3(n-2)+...+n.1/1^2+2^2+3^2+...+n^2

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Samyak Jain · Answer

Numerator of the limit is the summation (r = 1 to n) of r.(n – r + 1) = ∑ (nr – r 2 + r) = ∑ nr – ∑ r 2 + ∑ r = n∑r – ∑ r 2 + ∑r = n.n(n + 1)/2 – n(n + 1)(2n + 1)/6 + n(n + 1)/2 = [n(n + 1)/6][3n – 2n – 1 + 3] = n(n + 1)(n + 2)/6 Denominator is of the form ∑ (from r = 1 to n) r 2 = n(n + 1)(2n + 1)/6 lim n tends to {n(n + 1)(n + 2)/6} / {n(n + 1)(2n + 1)/6} = lim n tends to {(n + 1) / (2n + 1)} = 1/2 .

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limit x tends to infinity 1.n+2(n-1)+3(n-2)+...+n.1/1^2+2^2+3^2+...+n^2

limit x tends to infinity 1.n+2(n-1)+3(n-2)+...+n.1/1^2+2^2+3^2+...+n^2

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