Given the nth term for each sequence below, how to find the sum of the first 20 terms? (1) $a_{n} = 4 n + 1$ $a_n=4n+1$ (2) $a_{n} = n + \frac{1}{2}$ $a_n=n+1/2$

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Answer 1 · 2018-06-29T12:12:52

(1) 860 (2) 220

In both examples we will apply linearity over the sum and use:

$n \sum i = 1 i = \frac{n (n + 1)}{2}$ $sum_(i=1)^n i = (n(n+1))/2$ and $n \sum i = 1 1 = n$ $sum_(i=1)^n 1 = n$

(1) $a_{n} = 4 n + 1$ $a_n= 4n+1$

To find the sum of the first 20 terms:

$20 \sum i = 1 (4 i + 1) = 4 20 \sum i = 1 i + 20 \sum i = 1 1$ $sum_(i=1)^20 (4i+1) = 4sum_(i=1)^20 i+ sum_(i=1)^20 1$

$= 4 {\frac{20 (20 + 1)}{2}} + 20$ $= 4{{20(20+1))/2} + 20$

$= 4 \times 210 + 20 = 860$ $= 4xx210 +20 = 860$

(2) $a_{n} = n + \frac{1}{2}$ $a_n= n+1/2$

To find the sum of the first 20 terms:

$20 \sum i = 1 (i + \frac{1}{2}) = 20 \sum i = 1 i + \frac{1}{2} 20 \sum i = 1 1$ $sum_(i=1)^20 (i+1/2) = sum_(i=1)^20 i+ 1/2sum_(i=1)^20 1$

$= \frac{20 (20 + 1)}{2} + \frac{1}{2} \times 20$ $= (20(20+1))/2 + 1/2xx20$

$= 210 + 10 = 220$ $= 210 + 10 = 220$

Given the nth term for each sequence below, how to find the sum of the first 20 terms? (1) a_n=4n+1an=4n+1a_n=4n+1 (2) a_n=n+1/2an=n+12a_n=n+1/2