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Pani-rosa [81]
3 years ago
13

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

Mathematics
2 answers:
iragen [17]3 years ago
7 0

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

lana [24]3 years ago
3 0
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
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