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Lena [83]
3 years ago
7

Find a formula for the sum of n terms. Use the formula to find the limit as n → [infinity]. lim n → [infinity] n 2 + i n 8 n i =

1
Mathematics
1 answer:
Masteriza [31]3 years ago
5 0

Complete Question

Find a formula for the sum of n terms.   \sum\limits_{i=1}^n  ( 8 + \frac{i}{n} )(\frac{2}{n} )

Use the formula to find the limit as n \to \infty

 

Answer:

   K_n  =  \frac{n + 73 }{n}

  \lim_{n \to \infty} K_n  =  1

Step-by-step explanation:

     So let assume that

                  K_n  =  \sum\limits_{i=1}^n  ( 8 + \frac{i}{n} )(\frac{2}{n} )

=>             K_n  =  \sum\limits_{i=1}^n  ( \frac{16}{n} + \frac{2i}{n^2} )

=>              K_n  = \frac{2}{n}  \sum\limits_{i=1}^n (8) + \frac{2}{n^2}   \sum\limits_{i=1}^n(i)

Generally  

         \sum\limits_{i=1}^n (k) = \frac{1}{2}  n  (n + 1)

So  

      \sum\limits_{i=1}^n (8) = \frac{1}{2}  * 8*  (8 + 1)

      \sum\limits_{i=1}^n (8) = 36

K_n  = \frac{2}{n}  \sum\limits_{i=1}^n (8) + \frac{2}{n^2}   \sum\limits_{i=1}^n(i)  

and  

  \sum\limits_{i=1}^n (i) = \frac{1}{2}  n  (n + 1)

  Therefore

         K_n  = \frac{72}{n} + \frac{2}{n^2}   *  \frac{1}{2}  n (n + 1 )

         K_n  = \frac{72}{n} +    \frac{1}{n}   (n + 1 )

         K_n  = \frac{72}{n} +   1 +  \frac{1}{n}

        K_n  =  \frac{72 +  1 +  n }{n}

        K_n  =  \frac{n + 73 }{n}

Now  \lim_{n \to \infty} K_n  =  \lim_{n \to \infty} [\frac{n + 73 }{n} ]

=>     \lim_{n \to \infty} [\frac{n + 73 }{n} ]  =    \lim_{n \to \infty} [\frac{n}{n}  +  \frac{73 }{n}  ]

=>     \lim_{n \to \infty} [\frac{n + 73 }{n} ]  =    \lim_{n \to \infty} [1 +  \frac{73 }{n}  ]

=>     \lim_{n \to \infty} [\frac{n + 73 }{n} ]  =    \lim_{n \to \infty} [1 ] + \lim_{n \to \infty}  [\frac{73 }{n}  ]

=>    \lim_{n \to \infty} [\frac{n + 73 }{n} ]  =  1  +  0

Therefore

      \lim_{n \to \infty} K_n  =  1

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