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Question: L =\(\lim_{n \rightarrow \infty}\)<img src="https://cdn.pureessence.tech/canvas_662.png?top_left_x=0...

L =lim⁡n→∞\lim_{n \rightarrow \infty}

then eL is equal to –

A

1

B

1/7

C

7

D

e

Answer

7

Explanation

Solution

L = lim⁡n→∞\lim _ { n \rightarrow \infty } (n+2n2+n+1+n+4n2+2n+22+....+n+2(2n)n2+(2n)n+(2n)2)\left( \frac{n + 2}{n^{2} + n + 1} + \frac{n + 4}{n^{2} + 2n + 2^{2}} + .... + \frac{n + 2(2n)}{n^{2} + (2n)n + (2n)^{2}} \right)

= lim⁡n→∞\lim_{n \rightarrow \infty} ∑r=12nn+2rn2+rn+r2\sum_{r = 1}^{2n}\frac{n + 2r}{n^{2} + rn + r^{2}}

= lim⁡n→∞\lim_{n \rightarrow \infty}∑r=12n1+2(r/n)(r/n)2+(r/n)+1\sum_{r = 1}^{2n}\frac{1 + 2(r/n)}{(r/n)^{2} + (r/n) + 1}

= ∫021+2xx2+x+1\int_{0}^{2}\frac{1 + 2x}{x^{2} + x + 1}dx = [ln(x2+x+1)]02\left\lbrack \mathcal{l}n(x^{2} + x + 1) \right\rbrack_{0}^{2} = ln 7

\ eL = 7