Math, asked by vivek705, 1 year ago

lt n tends to infinity 1square +2square +3square....... +nsquare/nsquare(2n+1)

Answers

Answered by yogeshgupta20
47

please mark as brainlist

Attachments:
Answered by pulakmath007
1

\displaystyle  \sf\lim_{n \to  \infty } \:   \frac{ {1}^{2} +  {2}^{2}  +  {3}^{2}  + ... + {n}^{2}   }{ {n}^{2} (2n + 1)}= \frac{1}{6}

Given :

\displaystyle  \sf\lim_{n \to  \infty } \:   \frac{ {1}^{2} +  {2}^{2}  +  {3}^{2}  + ... + {n}^{2}   }{ {n}^{2} (2n + 1)}

To find : The limit value

Tip :

\displaystyle  \sf \:    {1}^{2} +  {2}^{2}  +  {3}^{2}  + ... + {n}^{2}=  \frac{n(n + 1)(2n + 1)}{6}

Solution :

\displaystyle  \sf\lim_{n \to  \infty } \:   \frac{ {1}^{2} +  {2}^{2}  +  {3}^{2}  + ... + {n}^{2}   }{ {n}^{2} (2n + 1)}

\displaystyle  \sf = \lim_{n \to  \infty } \:   \frac{ \frac{n(n + 1)(2n + 1)}{6} }{ {n}^{2} (2n + 1)}

\displaystyle  \sf = \lim_{n \to  \infty } \:   \frac{n(n + 1)(2n + 1)}{ 6{n}^{2} (2n + 1)}

\displaystyle  \sf = \lim_{n \to  \infty } \:   \frac{(n + 1)}{ 6n }

\displaystyle  \sf = \lim_{n \to  \infty } \:   \frac{(1 +  \frac{1}{n} )}{ 6 }

\displaystyle  \sf = \frac{1}{6}  \lim_{n \to  \infty } \:    \bigg(1 +  \frac{1}{n}  \bigg)

\displaystyle  \sf = \frac{1}{6}     \bigg(1  + 0 \bigg)

\displaystyle  \sf = \frac{1}{6}

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