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 [tex]a_n=2n+5[/tex], the sum of the first 15 terms is

[tex]\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[/tex]

Recall Faulhaber's formulas, which say

[tex]\displaystyle\sum_{n=1}^k1=k[/tex]
[tex]\displaystyle\sum_{n=1}^kn=\dfrac{k(k+1)}2[/tex]

Our sum is then

[tex]\displaystyle\sum_{n=1}^{15}a_n=2\dfrac{15\times16}2+5\times15=315[/tex]

Answer Link

athleticregina athleticregina · Answer 2 · 2019-02-27T11:45:01+01:00

Answer:

[tex]\sum _{n=1}^{15}2n+\sum _{n=1}^{15}5=240+75=315[/tex]

Step-by-step explanation:

Given: [tex]\sum _{n=1}^{15}\:2n+5[/tex]

We have to calculate the sum of given expression.

Consider the given expression [tex]\sum _{n=1}^{15}\:2n+5[/tex]

Apply sum rule,[tex]\sum a_n+b_n=\sum a_n+\sum b_n[/tex] , we get,

[tex]=\sum _{n=1}^{15}2n+\sum _{n=1}^{15}5[/tex]

Now first consider [tex]\sum _{n=1}^{15}2n[/tex]

Using constant multiplication rule, [tex]\sum c\cdot a_n=c\cdot \sum a_n[/tex]

we have,

[tex]=2\cdot \sum \:_{n=1}^{15}n[/tex]

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

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

Now consider [tex]\sum _{n=1}^{15}5[/tex]

Apply sum formula, [tex]\sum _{k=1}^n\:a\:=\:a\cdot n[/tex]

Here, a = 5 and n = 15

we get [tex]\sum _{n=1}^{15}5=75[/tex]

Therefore [tex]\sum _{n=1}^{15}2n+\sum _{n=1}^{15}5=240+75=315[/tex]