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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

Grade:12

1 Answers

Samyak Jain
333 Points
5 years ago
Numerator of the limit is the summation (r = 1 to n) of r.(n – r + 1) = ∑ (nr – r2 + r)
           = ∑ nr – ∑ r2 + ∑ r  =  n∑r – ∑ r2 + ∑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)  r2  =  n(n + 1)(2n + 1)/6
\therefore lim n tends to \infty {n(n + 1)(n + 2)/6} / {n(n + 1)(2n + 1)/6}  = lim n tends to \infty {(n + 1) / (2n + 1)}
                                         =  1/2 .

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