Find the sum of a finite arithmetic sequence from n = 1 to n = 15, using the expression 2n + 5

Assuming you mean the sequence is given explicitly by $a_n=2n+5$, the sum of the first 15 terms is

$\displaystyle\sum_{n=1}^{15}a_n=\sum_{n=1}^{15}(2n+5)=2\sum_{n=1}^{15}n+5\sum_{n=1}^{15}1$

Recall Faulhaber's formulas, which say

$\displaystyle\sum_{n=1}^k1=k$
$\displaystyle\sum_{n=1}^kn=\dfrac{k(k+1)}2$

Our sum is then

$\displaystyle\sum_{n=1}^{15}a_n=2\dfrac{15\times16}2+5\times15=315$

Answer Link

athleticregina athleticregina · Answer 2 · 2019-02-27T05:45:01-05:00

Answer:

$\sum _{n=1}^{15}2n+\sum _{n=1}^{15}5=240+75=315$

Step-by-step explanation:

Given: $\sum _{n=1}^{15}\:2n+5$

We have to calculate the sum of given expression.

Consider the given expression $\sum _{n=1}^{15}\:2n+5$

Apply sum rule,$\sum a_n+b_n=\sum a_n+\sum b_n$ , we get,

$=\sum _{n=1}^{15}2n+\sum _{n=1}^{15}5$

Now first consider $\sum _{n=1}^{15}2n$

Using constant multiplication rule, $\sum c\cdot a_n=c\cdot \sum a_n$

we have,

$=2\cdot \sum \:_{n=1}^{15}n$

Apply sum formula, $\sum _{k=1}^nk=\frac{1}{2}n\left(n+1\right)$

$=\frac{1}{2}\cdot \:15\left(15+1\right)=120\\\\ \sum _{n=1}^{15}2n=240$

Now consider $\sum _{n=1}^{15}5$

Apply sum formula, $\sum _{k=1}^n\:a\:=\:a\cdot n$

Here, a = 5 and n = 15

we get $\sum _{n=1}^{15}5=75$

Therefore $\sum _{n=1}^{15}2n+\sum _{n=1}^{15}5=240+75=315$