Answer :
Complete Question
Find a formula for the sum of n terms. [tex]\sum\limits_{i=1}^n ( 8 + \frac{i}{n} )(\frac{2}{n} )[/tex]
Use the formula to find the limit as [tex]n \to \infty[/tex]
Answer:
[tex]K_n = \frac{n + 73 }{n}[/tex]
[tex]\lim_{n \to \infty} K_n = 1[/tex]
Step-by-step explanation:
So let assume that
[tex]K_n = \sum\limits_{i=1}^n ( 8 + \frac{i}{n} )(\frac{2}{n} )[/tex]
=> [tex]K_n = \sum\limits_{i=1}^n ( \frac{16}{n} + \frac{2i}{n^2} )[/tex]
=> [tex]K_n = \frac{2}{n} \sum\limits_{i=1}^n (8) + \frac{2}{n^2} \sum\limits_{i=1}^n(i)[/tex]
Generally
[tex]\sum\limits_{i=1}^n (k) = \frac{1}{2} n (n + 1)[/tex]
So
[tex]\sum\limits_{i=1}^n (8) = \frac{1}{2} * 8* (8 + 1)[/tex]
[tex]\sum\limits_{i=1}^n (8) = 36[/tex]
[tex]K_n = \frac{2}{n} \sum\limits_{i=1}^n (8) + \frac{2}{n^2} \sum\limits_{i=1}^n(i)[/tex]
and
[tex]\sum\limits_{i=1}^n (i) = \frac{1}{2} n (n + 1)[/tex]
Therefore
[tex]K_n = \frac{72}{n} + \frac{2}{n^2} * \frac{1}{2} n (n + 1 )[/tex]
[tex]K_n = \frac{72}{n} + \frac{1}{n} (n + 1 )[/tex]
[tex]K_n = \frac{72}{n} + 1 + \frac{1}{n}[/tex]
[tex]K_n = \frac{72 + 1 + n }{n}[/tex]
[tex]K_n = \frac{n + 73 }{n}[/tex]
Now [tex]\lim_{n \to \infty} K_n = \lim_{n \to \infty} [\frac{n + 73 }{n} ][/tex]
=> [tex]\lim_{n \to \infty} [\frac{n + 73 }{n} ] = \lim_{n \to \infty} [\frac{n}{n} + \frac{73 }{n} ][/tex]
=> [tex]\lim_{n \to \infty} [\frac{n + 73 }{n} ] = \lim_{n \to \infty} [1 + \frac{73 }{n} ][/tex]
=> [tex]\lim_{n \to \infty} [\frac{n + 73 }{n} ] = \lim_{n \to \infty} [1 ] + \lim_{n \to \infty} [\frac{73 }{n} ][/tex]
=> [tex]\lim_{n \to \infty} [\frac{n + 73 }{n} ] = 1 + 0[/tex]
Therefore
[tex]\lim_{n \to \infty} K_n = 1[/tex]